How to find the largest value of the derivative of a function. How to find the largest and smallest values of a function in a bounded closed region? Main types of tasks
Let the function $z=f(x,y)$ be defined and continuous in some bounded closed domain $D$. Let the given function in this region have finite partial derivatives of the first order (except, perhaps, for a finite number of points). To find the largest and smallest values of a function of two variables in a given closed region, three steps of a simple algorithm are required.
Algorithm for finding the largest and smallest values of the function $z=f(x,y)$ in a closed domain $D$.
- Find the critical points of the function $z=f(x,y)$ belonging to the domain $D$. Calculate the function values at critical points.
- Investigate the behavior of the function $z=f(x,y)$ on the boundary of the region $D$, finding the points of possible maximum and minimum values. Calculate the function values at the obtained points.
- From the function values obtained in the previous two paragraphs, select the largest and smallest.
What are critical points? show\hide
Under critical points imply points at which both first-order partial derivatives are equal to zero (i.e. $\frac(\partial z)(\partial x)=0$ and $\frac(\partial z)(\partial y)=0 $) or at least one partial derivative does not exist.
Often the points at which first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset of critical points.
Example No. 1
Find the largest and smallest values of the function $z=x^2+2xy-y^2-4x$ in a closed region bounded by the lines $x=3$, $y=0$ and $y=x+1$.
We will follow the above, but first we will deal with the drawing of a given area, which we will denote by the letter $D$. We are given the equations of three straight lines that limit this area. The straight line $x=3$ passes through the point $(3;0)$ parallel to the ordinate axis (Oy axis). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct the line $y=x+1$, we will find two points through which we will draw this line. You can, of course, substitute a couple of arbitrary values instead of $x$. For example, substituting $x=10$, we get: $y=x+1=10+1=11$. We have found the point $(10;11)$ lying on the line $y=x+1$. However, it is better to find those points at which the straight line $y=x+1$ intersects the lines $x=3$ and $y=0$. Why is this better? Because we will kill a couple of birds with one stone: we will get two points to construct the straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines limiting the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ intersects at the point $(-1;0)$. In order not to clutter up the progress of the solution with auxiliary explanations, I will put the question of obtaining these two points in a note.
How were the points $(3;4)$ and $(-1;0)$ obtained? show\hide
Let's start from the intersection point of the lines $y=x+1$ and $x=3$. The coordinates of the desired point belong to both the first and second straight lines, therefore, to find the unknown coordinates, you need to solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$
The solution to such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.
Now let's find the intersection point of the lines $y=x+1$ and $y=0$. Let us again compose and solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$
Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (x-axis).
Everything is ready to build a drawing that will look like this:
The question of the note seems obvious, because everything can be seen from the picture. However, it is worth remembering that a drawing cannot serve as evidence. The drawing is for illustrative purposes only.
Our area was defined using the equations of lines that bound it. Obviously, these lines define a triangle, right? Or is it not entirely obvious? Or maybe we are given a different area, bounded by the same lines:
Of course, the condition says that the area is closed, so the picture shown is incorrect. But to avoid such ambiguities, it is better to define regions by inequalities. Are we interested in the part of the plane located under the straight line $y=x+1$? Ok, so $y ≤ x+1$. Should our area be located above the line $y=0$? Great, that means $y ≥ 0$. By the way, the last two inequalities can easily be combined into one: $0 ≤ y ≤ x+1$.
$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$
These inequalities define the region $D$, and they define it unambiguously, without allowing any ambiguity. But how does this help us with the question stated at the beginning of the note? It will also help :) We need to check whether the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:
$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right $$.
Both inequalities are valid. Point $M_1(1;1)$ belongs to region $D$.
Now it’s time to study the behavior of the function at the boundary of the region, i.e. let's go to . Let's start with the straight line $y=0$.
The straight line $y=0$ (x-axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Let's substitute $y=0$ into the given function $z(x,y)=x^2+2xy-y^2-4x$. We denote the function of one variable $x$ obtained as a result of substitution as $f_1(x)$:
$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$
Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of this function and equate it to zero:
$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$
The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we will also add $M_2(2;0)$ to the list of points. In addition, let us calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.
So, let's calculate the values of the function $z$ at points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points into the original expression $z=x^2+2xy-y^2-4x$. For example, for point $M_2$ we get:
$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$
However, the calculations can be simplified a little. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll write this down in detail:
\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)
Of course, there is usually no need for such detailed records, and in the future we will write down all calculations briefly:
$$z_2=f_1(2)=2^2-4\cdot 2=-4;\; z_3=f_1(-1)=(-1)^2-4\cdot (-1)=5;\; z_4=f_1(3)=3^2-4\cdot 3=-3.$$
Now let's turn to the straight line $x=3$. This straight line limits the region $D$ under the condition $0 ≤ y ≤ 4$. Let's substitute $x=3$ into the given function $z$. As a result of this substitution we get the function $f_2(y)$:
$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$
For the function $f_2(y)$ we need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Let's find the derivative of this function and equate it to zero:
$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$
The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we will also add $M_5(3;3)$ to the previously found points. In addition, you need to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at points $M_4(3;0)$ and $M_6(3;4)$. At point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:
\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; & z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)
And finally, consider the last boundary of the region $D$, i.e. straight line $y=x+1$. This straight line limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:
$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$
Once again we have a function of one variable $x$. And again we need to find the largest and smallest values of this function on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of the function $f_(3)(x)$ and equate it to zero:
$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$
The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. Points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we already found the value of the function in them.
$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$
The second step of the solution is completed. We received seven values:
$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$
Let's turn to . Choosing the largest and smallest values from the numbers obtained in the third paragraph, we will have:
$$z_(min)=-4; \; z_(max)=6.$$
The problem is solved, all that remains is to write down the answer.
Answer: $z_(min)=-4; \; z_(max)=6$.
Example No. 2
Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.
First, let's build a drawing. The equation $x^2+y^2=25$ (this is the boundary line of a given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ $25 satisfy all points inside and on the mentioned circle.
We will act according to. Let's find partial derivatives and find out the critical points.
$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$
There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. let's find stationary points.
$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8. \end(aligned) \right $$.
We have obtained a stationary point $(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check whether the inequality $x^2+y^2 ≤ 25$ holds, which defines our region $D$. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ does not hold. Conclusion: point $(6;-8)$ does not belong to area $D$.
So, there are no critical points inside the region $D$. Let's move on to... We need to study the behavior of a function on the boundary of a given region, i.e. on the circle $x^2+y^2=25$. We can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the equation of a circle we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:
$$ z=x^2+y^2-12x+16y=x^2+25-x^2-12x+16\sqrt(25-x^2)=25-12x+16\sqrt(25-x ^2); \;\; -5≤ x ≤ 5. $$
The further solution will be completely identical to the study of the behavior of the function at the boundary of the region in the previous example No. 1. However, it seems to me more reasonable to apply the Lagrange method in this situation. We will be interested only in the first part of this method. After applying the first part of the Lagrange method, we will obtain points at which we will examine the function $z$ for minimum and maximum values.
We compose the Lagrange function:
$$ F=z(x,y)+\lambda\cdot(x^2+y^2-25)=x^2+y^2-12x+16y+\lambda\cdot (x^2+y^2 -25). $$
We find the partial derivatives of the Lagrange function and compose the corresponding system of equations:
$$ F_(x)^(")=2x-12+2\lambda x; \;\; F_(y)^(")=2y+16+2\lambda y.\\ \left \( \begin (aligned) & 2x-12+2\lambda x=0;\\ & 2y+16+2\lambda y=0;\\ & x^2+y^2-25=0. right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$
To solve this system, let's immediately point out that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:
$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$
The resulting contradiction $0=6$ indicates that the value $\lambda=-1$ is unacceptable. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:
\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)
I believe that it becomes obvious here why we specifically stipulated the condition $\lambda\neq -1$. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator $1+\lambda\neq 0$.
Let us substitute the resulting expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:
$$ \left(\frac(6)(1+\lambda) \right)^2+\left(\frac(-8)(1+\lambda) \right)^2=25;\\ \frac( 36)((1+\lambda)^2)+\frac(64)((1+\lambda)^2)=25;\\ \frac(100)((1+\lambda)^2)=25 ; \; (1+\lambda)^2=4. $$
From the resulting equality it follows that $1+\lambda=2$ or $1+\lambda=-2$. Hence we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:
\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)
So, we have obtained two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Let's find the values of the function $z$ at points $M_1$ and $M_2$:
\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)
We should select the largest and smallest values from those we obtained in the first and second steps. But in this case the choice is small :) We have:
$$ z_(min)=-75; \; z_(max)=125. $$
Answer: $z_(min)=-75; \; z_(max)=$125.
Let the function y =f(X) is continuous on the interval [ a, b]. As is known, such a function reaches its maximum and minimum values on this segment. The function can take these values either at the internal point of the segment [ a, b], or on the boundary of the segment.
To find the largest and smallest values of a function on the segment [ a, b] necessary:
1) find the critical points of the function in the interval ( a, b);
2) calculate the values of the function at the found critical points;
3) calculate the values of the function at the ends of the segment, that is, when x=A and x = b;
4) from all calculated values of the function, select the largest and smallest.
Example. Find the largest and smallest values of a function
on the segment.
Finding critical points:
These points lie inside the segment ; y(1) = ‒ 3; y(2) = ‒ 4; y(0) = ‒ 8; y(3) = 1;
at the point x= 3 and at the point x= 0.
Study of a function for convexity and inflection point.
Function y = f (x) called convexup in between (a, b) , if its graph lies under the tangent drawn at any point in this interval, and is called convex down (concave), if its graph lies above the tangent.
The point through which convexity is replaced by concavity or vice versa is called inflection point.
Algorithm for examining convexity and inflection point:
1. Find critical points of the second kind, that is, points at which the second derivative is equal to zero or does not exist.
2. Plot critical points on the number line, dividing it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upward, if, then the function is convex downward.
3. If, when passing through a critical point of the second kind, the sign changes and at this point the second derivative is equal to zero, then this point is the abscissa of the inflection point. Find its ordinate.
Asymptotes of the graph of a function. Study of a function for asymptotes.
Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point on the graph to this line tends to zero as the point on the graph moves indefinitely from the origin.
There are three types of asymptotes: vertical, horizontal and inclined.
Definition. The straight line is called vertical asymptote function graphics y = f(x), if at least one of the one-sided limits of the function at this point is equal to infinity, that is
where is the discontinuity point of the function, that is, it does not belong to the domain of definition.
Example.
D ( y) = (‒ ∞; 2) (2; + ∞)
x= 2 – break point.
Definition. Straight y =A called horizontal asymptote function graphics y = f(x) at , if
Example.
|
x | |||
|
y |
Definition. Straight y =kx +b (k≠ 0) is called oblique asymptote function graphics y = f(x) at , where
General scheme for studying functions and constructing graphs.
Function Research Algorithmy = f(x) :
1. Find the domain of the function D (y).
2. Find (if possible) the points of intersection of the graph with the coordinate axes (if x= 0 and at y = 0).
3. Examine the evenness and oddness of the function ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).
4. Find the asymptotes of the graph of the function.
5. Find the intervals of monotonicity of the function.
6. Find the extrema of the function.
7. Find the intervals of convexity (concavity) and inflection points of the function graph.
8. Based on the research conducted, construct a graph of the function.
Example. Explore the function and build its graph.
1) D (y) =
x= 4 – break point.
2) When x = 0,
(0; ‒ 5) – point of intersection with oh.
At y = 0,
3)
y(‒
x)=
a function of general form (neither even nor odd).
4) We examine for asymptotes.
a) vertical
b) horizontal
c) find the oblique asymptotes where
‒oblique asymptote equation
5) In this equation it is not necessary to find intervals of monotonicity of the function.
6)
These critical points divide the entire domain of definition of the function into the interval (˗∞; ˗2), (˗2; 4), (4; 10) and (10; +∞). It is convenient to present the results obtained in the form of the following table.
Algorithm for finding the largest and smallest values of a continuous function on a segment:
1) Find all critical points of the function belonging to the segment ;
2) Calculate the values of the function at these points and at the ends of the segment;
3) From the obtained values, select the largest and smallest.
Example 8.1. Find the largest and smallest values of a function 
on the segment 
.
Solution. 1) Find the critical points of the function.
,
.
On the segment 
the denominator does not vanish. Therefore, a fraction is equal to zero if and only if the numerator is equal to zero:
.
Means, 
– critical point of the function. It belongs to this segment.
Let's find the value of the function at the critical point:
2) Find the values of the function at the ends of the segment:
,
.
3) From the obtained values, select the largest and smallest:
, 
.
9. Problems of finding the largest and smallest values of quantities
When solving problems involving calculating the smallest and largest values of quantities, you must first determine for which quantity in the problem you need to find the smallest or largest value. This value will be the function under study. Then one of the quantities on the change of which the application of the function depends should be taken as an independent variable and the function expressed through it. In this case, it is necessary to choose as the independent variable the value through which the function under study is expressed most simply. After this, the problem of finding the smallest and largest values of the resulting function in a certain interval of change in the independent variable is solved, which is usually established from the very essence of the problem.
Example 9.1. Find the height of the cone of the largest volume that can be inscribed in a ball of radius 
.
R 
decision. Designating the radius of the base, height and volume of the cone, respectively 
,
And 
, let's write 
. This equality expresses the dependence on two variables 
And 
; let us exclude one of these quantities, namely 
. To do this, from a right triangle 
we derive (using the theorem about the square of a perpendicular dropped from the vertex of a right angle to the hypotenuse):
Figure 6 – Illustration for example 9.1.
or
.
Substituting the value 
into the formula for the volume of a cone, we get:
.
We see that the volume 
cone inscribed in a ball of radius 
, there is a function of the height of this cone 
. Finding the height at which the inscribed cone has a large volume means finding such 
, at which the function 
has a maximum. We are looking for the maximum function:
1)
,
2)
,
,
, where 
or 
,
3)
.
Substituting instead 
at first 
, and then 
, we get:
In the first case we have a minimum ( 
at 
), in the second the desired maximum (since 
at 
).
Therefore, when 
cone inscribed in a ball of radius 
,
has the largest volume.
P 
Example 9.2.
It is required to fence with a wire mesh length 60
m a rectangular area adjacent to the wall of the house (Fig. 7). What should be the length and width of the plot so that it has the largest area?
Solution. Let the width of the plot 
m, and the area 
m 2
, Then:
Figure 7 – Illustration for example 9.2.
Values 
And 
cannot be negative, so the multiplier
, A 
.
Square 
there is a function 
, we determine the intervals of its increase and decrease:
.
, and the function increases when 
;
, and the function decreases when 
. Therefore, the point 
is the maximum point. Since this is the only point belonging to the interval 
, then at the point 
function matters most.
Therefore, the area of the plot is greatest (maximum) if the width 
m, and the length
m.
Example 9.3. What should be the dimensions of a rectangular room whose area 36 m 2 so that its perimeter is smallest?
Solution. Let the length be 
m, then the width of the rectangle 
m, and the perimeter:
.
Perimeter 
there is a function of length 
, defined for all positive values 
:
.
Let us determine the intervals of its increase and decrease:
The sign of the derivative is determined by the sign of the difference 
. In between
, and in between 
.
Therefore, the point 
is the minimum point. Since this is the only point belonging to the interval: 
, then at the point 
function has the smallest value.
Therefore, the perimeter of a rectangle has the smallest value (minimum) if its length 6
m and width 
m = 6 m, that is, when it is a square.
The largest (smallest) value of a function is the largest (smallest) accepted value of the ordinate on the considered interval.
To find the largest or smallest value of a function you need to:
- Check which stationary points are included in a given segment.
- Calculate the value of the function at the ends of the segment and at stationary points from step 3
- Select the largest or smallest value from the results obtained.
To find the maximum or minimum points you need to:
- Find the derivative of the function $f"(x)$
- Find stationary points by solving the equation $f"(x)=0$
- Factor the derivative of a function.
- Draw a coordinate line, place stationary points on it and determine the signs of the derivative in the resulting intervals, using the notation in step 3.
- Find the maximum or minimum points according to the rule: if at a point the derivative changes sign from plus to minus, then this will be the maximum point (if from minus to plus, then this will be the minimum point). In practice, it is convenient to use the image of arrows on intervals: on the interval where the derivative is positive, the arrow is drawn upward and vice versa.
Table of derivatives of some elementary functions:
| Function | Derivative |
| $c$ | $0$ |
| $x$ | $1$ |
| $x^n, n∈N$ | $nx^(n-1), n∈N$ |
| $(1)/(x)$ | $-(1)/(x^2)$ |
| $(1)/x(^n), n∈N$ | $-(n)/(x^(n+1)), n∈N$ |
| $√^n(x), n∈N$ | $(1)/(n√^n(x^(n-1)), n∈N$ |
| $sinx$ | $cosx$ |
| $cosx$ | $-sinx$ |
| $tgx$ | $(1)/(cos^2x)$ |
| $ctgx$ | $-(1)/(sin^2x)$ |
| $cos^2x$ | $-sin2x$ |
| $sin^2x$ | $sin2x$ |
| $e^x$ | $e^x$ |
| $a^x$ | $a^xlna$ |
| $lnx$ | $(1)/(x)$ |
| $log_(a)x$ | $(1)/(xlna)$ |
Basic rules of differentiation
1. The derivative of the sum and difference is equal to the derivative of each term
$(f(x) ± g(x))′= f′(x)± g′(x)$
Find the derivative of the function $f(x) = 3x^5 – cosx + (1)/(x)$
The derivative of the sum and difference is equal to the derivative of each term
$f′(x)=(3x^5)′–(cosx)′+((1)/(x))"=15x^4+sinx-(1)/(x^2)$
2. Derivative of the product.
$(f(x)∙g(x))′=f′(x)∙g(x)+f(x)∙g(x)′$
Find the derivative $f(x)=4x∙cosx$
$f′(x)=(4x)′∙cosx+4x∙(cosx)′=4∙cosx-4x∙sinx$
3. Derivative of the quotient
$((f(x))/(g(x)))"=(f^"(x)∙g(x)-f(x)∙g(x)")/(g^2(x) )$
Find the derivative $f(x)=(5x^5)/(e^x)$
$f"(x)=((5x^5)"∙e^x-5x^5∙(e^x)")/((e^x)^2)=(25x^4∙e^x- 5x^5∙e^x)/((e^x)^2)$
4. The derivative of a complex function is equal to the product of the derivative of the external function and the derivative of the internal function
$f(g(x))′=f′(g(x))∙g′(x)$
$f′(x)=cos′(5x)∙(5x)′= - sin(5x)∙5= -5sin(5x)$
Find the minimum point of the function $y=2x-ln(x+11)+4$
1. Find the ODZ of the function: $x+11>0; x>-11$
2. Find the derivative of the function $y"=2-(1)/(x+11)=(2x+22-1)/(x+11)=(2x+21)/(x+11)$
3. Find stationary points by equating the derivative to zero
$(2x+21)/(x+11)=0$
A fraction is equal to zero if the numerator is zero and the denominator is not zero.
$2x+21=0; x≠-11$
4. Let's draw a coordinate line, place stationary points on it and determine the signs of the derivative in the resulting intervals. To do this, substitute any number from the rightmost region into the derivative, for example, zero.
$y"(0)=(2∙0+21)/(0+11)=(21)/(11)>0$
5. At the minimum point, the derivative changes sign from minus to plus, therefore, the point $-10.5$ is the minimum point.
Answer: $-10.5$
Find the greatest value of the function $y=6x^5-90x^3-5$ on the segment $[-5;1]$
1. Find the derivative of the function $y′=30x^4-270x^2$
2. Equate the derivative to zero and find stationary points
$30x^4-270x^2=0$
Let's take the total factor $30x^2$ out of brackets
$30x^2(x^2-9)=0$
$30x^2(x-3)(x+3)=0$
Let's equate each factor to zero
$x^2=0 ; x-3=0; x+3=0$
$x=0;x=3;x=-3$
3. Select stationary points that belong to the given segment $[-5;1]$
The stationary points $x=0$ and $x=-3$ suit us
4. Calculate the value of the function at the ends of the segment and at stationary points from step 3
In practice, it is quite common to use the derivative in order to calculate the largest and smallest value of a function. We perform this action when we figure out how to minimize costs, increase profits, calculate the optimal load on production, etc., that is, in cases where we need to determine the optimal value of a parameter. To solve such problems correctly, you need to have a good understanding of what the largest and smallest values of a function are.
Typically we define these values within a certain interval x, which in turn may correspond to the entire domain of the function or part of it. It can be like a segment [a; b ] , and open interval (a ; b), (a ; b ], [ a ; b), infinite interval (a ; b), (a ; b ], [ a ; b) or infinite interval - ∞ ; a , (- ∞ ; a ] , [ a ; + ∞) , (- ∞ ; + ∞) .
In this material we will tell you how to calculate the largest and smallest values of an explicitly defined function with one variable y=f(x) y = f (x) .
Basic definitions
Let's start, as always, with the formulation of basic definitions.
Definition 1
The largest value of the function y = f (x) on a certain interval x is the value m a x y = f (x 0) x ∈ X, which for any value x x ∈ X, x ≠ x 0 makes the inequality f (x) ≤ f (x) valid 0) .
Definition 2
The smallest value of the function y = f (x) on a certain interval x is the value m i n x ∈ X y = f (x 0) , which for any value x ∈ X, x ≠ x 0 makes the inequality f(X f (x) ≥ f (x 0) .
These definitions are quite obvious. Even simpler, we can say this: the largest value of a function is its largest value on a known interval at abscissa x 0, and the smallest is the smallest accepted value on the same interval at x 0.
Definition 3
Stationary points are those values of the argument of a function at which its derivative becomes 0.
Why do we need to know what stationary points are? To answer this question, we need to remember Fermat's theorem. It follows from it that a stationary point is the point at which the extremum of the differentiable function is located (i.e., its local minimum or maximum). Consequently, the function will take the smallest or largest value on a certain interval precisely at one of the stationary points.
A function can also take on the largest or smallest value at those points at which the function itself is defined and its first derivative does not exist.
The first question that arises when studying this topic: in all cases can we determine the largest or smallest value of a function on a given interval? No, we cannot do this when the boundaries of a given interval coincide with the boundaries of the definition area, or if we are dealing with an infinite interval. It also happens that a function in a given segment or at infinity will take infinitely small or infinitely large values. In these cases, it is not possible to determine the largest and/or smallest value.
These points will become clearer after being depicted on the graphs:
The first figure shows us a function that takes the largest and smallest values (m a x y and m i n y) at stationary points located on the segment [ - 6 ; 6].
Let us examine in detail the case indicated in the second graph. Let's change the value of the segment to [ 1 ; 6 ] and we find that the maximum value of the function will be achieved at the point with the abscissa at the right boundary of the interval, and the minimum - at the stationary point.
In the third figure, the abscissas of the points represent the boundary points of the segment [ - 3 ; 2]. They correspond to the largest and smallest value of a given function.
Now let's look at the fourth picture. In it, the function takes m a x y (the largest value) and m i n y (the smallest value) at stationary points on the open interval (- 6 ; 6).
If we take the interval [ 1 ; 6), then we can say that the smallest value of the function on it will be achieved at a stationary point. The greatest value will be unknown to us. The function could take its maximum value at x equal to 6 if x = 6 belonged to the interval. This is exactly the case shown in graph 5.
In graph 6, this function acquires its smallest value at the right boundary of the interval (- 3; 2 ], and we cannot draw definite conclusions about the largest value.
In Figure 7 we see that the function will have m a x y at a stationary point having an abscissa equal to 1. The function will reach its minimum value at the boundary of the interval on the right side. At minus infinity, the function values will asymptotically approach y = 3.
If we take the interval x ∈ 2; + ∞ , then we will see that the given function will take neither the smallest nor the largest value on it. If x tends to 2, then the values of the function will tend to minus infinity, since the straight line x = 2 is a vertical asymptote. If the abscissa tends to plus infinity, then the function values will asymptotically approach y = 3. This is exactly the case shown in Figure 8.
In this paragraph we will present the sequence of actions that need to be performed to find the largest or smallest value of a function on a certain segment.
- First, let's find the domain of definition of the function. Let's check whether the segment specified in the condition is included in it.
- Now let's calculate the points contained in this segment at which the first derivative does not exist. Most often they can be found in functions whose argument is written under the modulus sign, or in power functions whose exponent is a fractionally rational number.
- Next, we will find out which stationary points will fall in the given segment. To do this, you need to calculate the derivative of the function, then equate it to 0 and solve the resulting equation, and then select the appropriate roots. If we don’t get a single stationary point or they don’t fall into the given segment, then we move on to the next step.
- We determine what values the function will take at given stationary points (if any), or at those points at which the first derivative does not exist (if there are any), or we calculate the values for x = a and x = b.
- 5. We have a number of function values, from which we now need to select the largest and smallest. These will be the largest and smallest values of the function that we need to find.
Let's see how to correctly apply this algorithm when solving problems.
Example 1
Condition: the function y = x 3 + 4 x 2 is given. Determine its largest and smallest values on the segments [ 1 ; 4 ] and [ - 4 ; - 1 ] .
Solution:
Let's start by finding the domain of definition of a given function. In this case, it will be the set of all real numbers except 0. In other words, D (y) : x ∈ (- ∞ ; 0) ∪ 0 ; + ∞ . Both segments specified in the condition will be inside the definition area.
Now we calculate the derivative of the function according to the rule of fraction differentiation:
y " = x 3 + 4 x 2 " = x 3 + 4 " x 2 - x 3 + 4 x 2 " x 4 = = 3 x 2 x 2 - (x 3 - 4) 2 x x 4 = x 3 - 8 x 3
We learned that the derivative of a function will exist at all points of the segments [ 1 ; 4 ] and [ - 4 ; - 1 ] .
Now we need to determine the stationary points of the function. Let's do this using the equation x 3 - 8 x 3 = 0. It has only one real root, which is 2. It will be a stationary point of the function and will fall into the first segment [1; 4].
Let us calculate the values of the function at the ends of the first segment and at this point, i.e. for x = 1, x = 2 and x = 4:
y (1) = 1 3 + 4 1 2 = 5 y (2) = 2 3 + 4 2 2 = 3 y (4) = 4 3 + 4 4 2 = 4 1 4
We found that the largest value of the function m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 will be achieved at x = 1, and the smallest m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 – at x = 2.
The second segment does not include a single stationary point, so we need to calculate the function values only at the ends of the given segment:
y (- 1) = (- 1) 3 + 4 (- 1) 2 = 3
This means m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .
Answer: For the segment [ 1 ; 4 ] - m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 , m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 , for the segment [ - 4 ; - 1 ] - m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .
See picture:
Before studying this method, we advise you to review how to correctly calculate the one-sided limit and the limit at infinity, as well as learn the basic methods for finding them. To find the largest and/or smallest value of a function on an open or infinite interval, perform the following steps sequentially.
- First, you need to check whether the given interval will be a subset of the domain of the given function.
- Let us determine all points that are contained in the required interval and at which the first derivative does not exist. They usually occur in functions where the argument is enclosed in the modulus sign, and in power functions with a fractionally rational exponent. If these points are missing, then you can proceed to the next step.
- Now let’s determine which stationary points will fall within the given interval. First, we equate the derivative to 0, solve the equation and select suitable roots. If we do not have a single stationary point or they do not fall within the specified interval, then we immediately proceed to further actions. They are determined by the type of interval.
- If the interval is of the form [ a ; b) , then we need to calculate the value of the function at the point x = a and the one-sided limit lim x → b - 0 f (x) .
- If the interval has the form (a; b ], then we need to calculate the value of the function at the point x = b and the one-sided limit lim x → a + 0 f (x).
- If the interval has the form (a; b), then we need to calculate the one-sided limits lim x → b - 0 f (x), lim x → a + 0 f (x).
- If the interval is of the form [ a ; + ∞), then we need to calculate the value at the point x = a and the limit at plus infinity lim x → + ∞ f (x) .
- If the interval looks like (- ∞ ; b ] , we calculate the value at the point x = b and the limit at minus infinity lim x → - ∞ f (x) .
- If - ∞ ; b , then we consider the one-sided limit lim x → b - 0 f (x) and the limit at minus infinity lim x → - ∞ f (x)
- If - ∞; + ∞ , then we consider the limits on minus and plus infinity lim x → + ∞ f (x) , lim x → - ∞ f (x).
- At the end, you need to draw a conclusion based on the obtained function values and limits. There are many options available here. So, if the one-sided limit is equal to minus infinity or plus infinity, then it is immediately clear that nothing can be said about the smallest and largest values of the function. Below we will look at one typical example. Detailed descriptions will help you understand what's what. If necessary, you can return to Figures 4 - 8 in the first part of the material.
Condition: given function y = 3 e 1 x 2 + x - 6 - 4 . Calculate its largest and smallest value in the intervals - ∞ ; - 4, - ∞; - 3 , (- 3 ; 1 ] , (- 3 ; 2) , [ 1 ; 2) , 2 ; + ∞ , [ 4 ; + ∞) .
Solution
First of all, we find the domain of definition of the function. The denominator of the fraction contains a quadratic trinomial, which should not turn to 0:
x 2 + x - 6 = 0 D = 1 2 - 4 1 (- 6) = 25 x 1 = - 1 - 5 2 = - 3 x 2 = - 1 + 5 2 = 2 ⇒ D (y) : x ∈ (- ∞ ; - 3) ∪ (- 3 ; 2) ∪ (2 ; + ∞)
We have obtained the domain of definition of the function to which all the intervals specified in the condition belong.
Now let's differentiate the function and get:
y" = 3 e 1 x 2 + x - 6 - 4 " = 3 e 1 x 2 + x - 6 " = 3 e 1 x 2 + x - 6 1 x 2 + x - 6 " = = 3 · e 1 x 2 + x - 6 · 1 " · x 2 + x - 6 - 1 · x 2 + x - 6 " (x 2 + x - 6) 2 = - 3 · (2 x + 1) · e 1 x 2 + x - 6 x 2 + x - 6 2
Consequently, derivatives of a function exist throughout its entire domain of definition.
Let's move on to finding stationary points. The derivative of the function becomes 0 at x = - 1 2 . This is a stationary point that lies in the intervals (- 3 ; 1 ] and (- 3 ; 2) .
Let's calculate the value of the function at x = - 4 for the interval (- ∞ ; - 4 ], as well as the limit at minus infinity:
y (- 4) = 3 e 1 (- 4) 2 + (- 4) - 6 - 4 = 3 e 1 6 - 4 ≈ - 0 . 456 lim x → - ∞ 3 e 1 x 2 + x - 6 = 3 e 0 - 4 = - 1
Since 3 e 1 6 - 4 > - 1, it means that m a x y x ∈ (- ∞ ; - 4 ] = y (- 4) = 3 e 1 6 - 4. This does not allow us to uniquely determine the smallest value of the function. We can only conclude that there is a constraint below - 1, since it is to this value that the function approaches asymptotically at minus infinity.
The peculiarity of the second interval is that there is not a single stationary point and not a single strict boundary in it. Consequently, we will not be able to calculate either the largest or smallest value of the function. Having defined the limit at minus infinity and as the argument tends to - 3 on the left side, we get only an interval of values:
lim x → - 3 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 - 0 3 e 1 (x + 3) (x - 3) - 4 = 3 e 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → - ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
This means that the function values will be located in the interval - 1; +∞
To find the largest value of the function in the third interval, we determine its value at the stationary point x = - 1 2 if x = 1. We will also need to know the one-sided limit for the case when the argument tends to - 3 on the right side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e 4 25 - 4 ≈ - 1 . 444 y (1) = 3 e 1 1 2 + 1 - 6 - 4 ≈ - 1 . 644 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 - 3 + 0 + 3 (- 3 + 0 - 2) - 4 = = 3 e 1 (- 0) - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4
It turned out that the function will take the greatest value at a stationary point m a x y x ∈ (3; 1 ] = y - 1 2 = 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. Everything we know , is the presence of a lower limit to - 4 .
For the interval (- 3 ; 2), take the results of the previous calculation and once again calculate what the one-sided limit is equal to when tending to 2 on the left side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e - 4 25 - 4 ≈ - 1 . 444 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = - 4 lim x → 2 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 - 0 + 3) (2 - 0 - 2) - 4 = = 3 e 1 - 0 - 4 = 3 e - ∞ - 4 = 3 · 0 - 4 = - 4
This means that m a x y x ∈ (- 3 ; 2) = y - 1 2 = 3 e - 4 25 - 4, and the smallest value cannot be determined, and the values of the function are limited from below by the number - 4.
Based on what we got in the two previous calculations, we can say that on the interval [ 1 ; 2) the function will take the largest value at x = 1, but it is impossible to find the smallest.
On the interval (2 ; + ∞) the function will not reach either the largest or the smallest value, i.e. it will take values from the interval - 1 ; + ∞ .
lim x → 2 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → + ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
Having calculated what the value of the function will be equal to at x = 4, we find out that m a x y x ∈ [ 4 ; + ∞) = y (4) = 3 e 1 14 - 4 , and the given function at plus infinity will asymptotically approach the straight line y = - 1 .
Let's compare what we got in each calculation with the graph of the given function. In the figure, the asymptotes are shown by dotted lines.
That's all we wanted to tell you about finding the largest and smallest values of a function. The sequences of actions that we have given will help you make the necessary calculations as quickly and simply as possible. But remember that it is often useful to first find out at which intervals the function will decrease and at which it will increase, after which you can draw further conclusions. This way you can more accurately determine the largest and smallest values of the function and justify the results obtained.
If you notice an error in the text, please highlight it and press Ctrl+Enter
