# Calculus

## Applications of the Derivative # Local Extrema of Functions

• ### Definition of Local Maximum and Local Minimum

Let a function $$y = f\left( x \right)$$ be defined in a $$\delta$$-neighborhood of a point $${x_0},$$ where $$\delta > 0.$$ The function $$f\left( x \right)$$ is said to have a local (or relative) maximum at the point $${x_0},$$ if for all points $$x \ne {x_0}$$ belonging to the neighborhood $$\left( {{x_0} – \delta ,{x_0} + \delta } \right)$$ the following inequality holds:

$f\left( x \right) \le f\left( {{x_0}} \right).$

If the strict inequality holds for all points $$x \ne {x_0}$$ in some neighborhood of $${x_0}:$$

$f\left( x \right) \lt f\left( {{x_0}} \right),$

then the point $${x_0}$$ is a strict local maximum point.

Similarly, we define a local (or relative) minimum of the function $$f\left( x \right).$$ In this case, the following inequality is valid for all points $$x \ne {x_0}$$ of the $$\delta$$-neighborhood $$\left( {{x_0} – \delta ,{x_0} + \delta } \right)$$ of the point $${x_0}:$$

$f\left( x \right) \ge f\left( {{x_0}} \right).$

Accordingly, a strict local minimum is described by the inequality

$f\left( x \right) \gt f\left( {{x_0}} \right).$

The concepts of local maximum and local minimum are united under the general term local extremum. The word “local” is often ommitted for brevity, so it is said simply about maxima and minima of functions.

Figure $$1$$ schematically shows the different extrema points. The point $$A\left( {{x_1}} \right)$$ is a strict local minimum point, since there exists a $$\delta$$-neighborhood $$\left( {{x_1} – \delta ,{x_1} + \delta } \right),$$ in which the following inequality holds:

${f\left( x \right) > f\left( {{x_1}} \right)\;}\kern0pt {\forall\;x \in \left( {{x_1} – \delta ,{x_1} + \delta } \right).}$

Similarly, the point $$B\left( {{x_2}} \right)$$ is a strict local maximum point. At this point, we have the inequality

${f\left( x \right) < f\left( {{x_2}} \right)\;}\kern0pt {\forall\;x \in \left( {{x_2} – \delta ,{x_2} + \delta } \right).}$

(Of course, the number $$\delta$$ at each point may be different.)

The subsequent points are classified as follows:

• $$C\left( {{x_3}} \right)$$ is a strict minimum point;
• $$D\left( {{x_4}} \right)$$ is a non-strict maximum point;
• $$E\left( {{x_5}} \right)$$ is a non-strict maximum or minimum point;
• $$F\left( {{x_6}} \right)$$ is a non-strict maximum point;
• $$G\left( {{x_7}} \right)$$ is a non-strict minimum point;
• $$H\left( {{x_8}} \right)$$ is a non-strict maximum or minimum point;
• $$I\left( {{x_9}} \right)$$ is a non-strict maximum point;
• $$J\left( {{x_{10}}} \right)$$ − there is no extremum.

### Necessary Condition for an Extremum

We introduce some more concepts.

The points at which the derivative of the function $$f\left( x \right)$$ is equal to zero are called the stationary points.

The points at which the derivative of the function $$f\left( x \right)$$ is equal to zero or does not exist are called the critical points of the function. Consequently, the stationary points are a subset of the set of critical points.

A necessary condition for an extremum is formulated as follows:

If the point $${x_0}$$ is an extremum point of the function $$f\left( x \right),$$ then the derivative at this point either is zero or does not exist. In other words, the extrema of a function are contained among its critical points.

The proof of the necessary condition follows from Fermat’s theorem.

Note that the necessary condition does not guarantee the existence of an extremum. A classic illustration here is the cubic function $$f\left( x \right) = {x^3}.$$ Despite the fact that the derivative of the function at the point $$x = 0$$ is zero: $$f’\left( {x = 0} \right) = 0,$$ this point is not an extremum.

Local extrema of differentiable functions exist when the sufficient conditions are satisfied. These conditions are based on the use of the first-, second-, or higher-order derivative. Respectively, $$3$$ sufficient conditions for local extrema are considered. Now we turn to their formulation and proof.

### First Derivative Test

Let the function $$f\left( x \right)$$ be differentiable in a neighborhood of the point $${x_0},$$ except perhaps at the point $${x_0}$$ itself, in which, however, the function is continuous. Then:

• If the derivative $$f’\left( x \right)$$ changes sign from minus to plus when passing through the point $${x_0}$$ (from left to right), then $${x_0}$$ is a strict minimum point (Figure $$2$$). In other words, in this case there exists a number $$\delta \gt 0$$ such that
${\forall \;x \in \left( {{x_0} – \delta ,{x_0}} \right)} \Rightarrow {f’\left( x \right) \lt 0,}$
${\forall \;x \in \left( {{x_0}, {x_0} + \delta} \right)} \Rightarrow {f’\left( x \right) \gt 0.}$
• If the derivative $$f’\left( x \right),$$ on the contrary, changes sign from plus to minus when passing through the point $${x_0},$$ then $${x_0}$$ is a strict maximum point (Figure $$3$$). In other words, there exists a number $$\delta > 0$$ such that
${\forall \;x \in \left( {{x_0} – \delta ,{x_0}} \right)} \Rightarrow {f’\left( x \right) \gt 0,}$
${\forall \;x \in \left( {{x_0}, {x_0} + \delta} \right)} \Rightarrow {f’\left( x \right) \lt 0.}$

#### Proof.

We confine ourselves to the case of the minimum. Suppose that the derivative $$f’\left( x \right)$$ changes sign from minus to plus when passing through the point $${x_0}.$$ To the left from the point $${x_0},$$ the following condition is satisfied:

${\forall \;x \in \left( {{x_0} – \delta ,{x_0}} \right)} \Rightarrow {f’\left( x \right) < 0.}$

By Lagrange’s theorem, the difference of the values of the function at the points $$x$$ and $${x_0}$$ is written as

${f\left( x \right) – f\left( {{x_0}} \right)} = {f’\left( c \right)\left( {x – \;{x_0}} \right),}$

where the point $$c$$ belongs to the interval $$\left( {{x_0} – \delta ,{x_0}} \right),$$ in which the derivative is negative, i.e. $$f’\left( c \right) < 0.$$ Since $$x – {x_0} < 0$$ to the left of the point $${x_0},$$ then

${f\left( x \right) – f\left( {{x_0}} \right) > 0}\;\;\kern-0.3pt{\text{for all}\;\;x \in \left( {{x_0} – \delta ,{x_0}} \right).}$

Likewise, it is established that

${f\left( x \right) – f\left( {{x_0}} \right) > 0}\;\;\kern-0.3pt{\text{for all}\;\;x \in \left( {{x_0}, {x_0} + \delta} \right).}$

(to the right of the point $${x_0}$$).

Based on the definition, we conclude that $${x_0}$$ is a strict minimum point of the function $$f\left( x \right).$$

Similarly, we can prove the first derivative test for a strict maximum.

Note that the first derivative test does not require the function to be differentiable at the point $${x_0}.$$ If the derivative at this point is infinite or does not exist (i.e. the point $${x_0}$$ is critical, but not stationary), the first derivative test can still be used to investigate the local extrema of the function.

### Second Derivative Test

Let the first derivative of a function $$f\left( x \right)$$ at the point$${x_0}$$ be equal to zero: $$f\left( {{x_0}} \right) = 0,$$ that is $${x_0}$$ is a stationary point of $$f\left( x \right).$$ Suppose also that there exists the second derivative $$f^{\prime\prime}\left( {{x_0}} \right)$$ at this point. Then

• If $$f^{\prime\prime}\left( {{x_0}} \right) \gt 0,$$ then $${x_0}$$ is a strict minimum point of the function $$f\left( x \right)$$;
• If $$f^{\prime\prime}\left( {{x_0}} \right) \lt 0,$$ then $${x_0}$$ is a strict maximum point of the function $$f\left( x \right).$$

#### Proof.

In the case of a strict minimum $$f^{\prime\prime}\left( {{x_0}} \right) \gt 0.$$ Then the first derivative is an increasing function at the point $${x_0}.$$ Consequently, there exists a number $$\delta \gt 0$$ such that

${\forall \;x \in \left( {{x_0} – \delta ,{x_0}} \right)} \Rightarrow {f’\left( x \right) \lt f’\left( {{x_0}} \right),}$

${\forall \;x \in \left( {{x_0}, {x_0} + \delta} \right)} \Rightarrow {f’\left( x \right) \gt f’\left( {{x_0}} \right).}$

Since $$f^{\prime\prime}\left( {{x_0}} \right) = 0$$ (because $${x_0}$$ is a stationary point), therefore the first derivative is negative in the $$\delta$$-neighborhood to the left of the point $${x_0}$$, and is positive to the right, i.e. the derivative changes sign from minus to plus when passing through the point $${x_0}.$$ By the first derivative test, this means that $${x_0}$$ is a strict minimum point.

The case of the maximum can be considered in a similar way.

The second derivative test is convenient to use when calculation of the first derivatives in the neighborhood of a stationary point is difficult. On the other hand, the second test may be used only for stationary points (where the first derivative is zero) − in contrast to the first derivative test, which is applicable to any critical points.

### Third Derivative Test

Let the function $$f\left( x \right)$$ have derivatives at the point $${x_0}$$ up to the $$n$$th order inclusively. Then if

${f’\left( {{x_0}} \right) } = {f^{\prime\prime}\left( {{x_0}} \right) = \ldots } = {{f^{\left( {n – 1} \right)}}\left( {{x_0}} \right) = 0\;\;\;}\kern-0.3pt {\text{and}\;\;{f^{\left( n \right)}}\left( {{x_0}} \right) \ne 0,}$

the point $${x_0}$$ for even $$n$$ is

• a strict minimum point if $${f^{\left( n \right)}}\left( {{x_0}} \right) \gt 0,$$ and
• a strict maximum point if $${f^{\left( n \right)}}\left( {{x_0}} \right) \gt 0.$$

For odd $$n,$$ the extremum at $${x_0}$$ does not exist.

It is clear that for $$n = 2$$, we obtain as a special case the second derivative test for local extrema considered above. To avoid such a transition, the third derivative test implies that $$n \gt 2.$$

#### Proof.

Expand the function $$f\left( x \right)$$ at the point $${x_0}$$ in a Taylor series:

${f\left( x \right) = f\left( {{x_0}} \right) } + {\frac{{f’\left( {{x_0}} \right)}}{{1!}}\left( {x – {x_0}} \right) } + {\frac{{f^{\prime\prime}\left( {{x_0}} \right)}}{{2!}}{\left( {x – {x_0}} \right)^2} + \ldots } + {\frac{{{f^{\left( {n – 1} \right)}}\left( {{x_0}} \right)}}{{\left( {n – 1} \right)!}}{\left( {x – {x_0}} \right)^{n – 1}} } + {\frac{{{f^{\left( n \right)}}\left( {{x_0}} \right)}}{{n!}}{\left( {x – {x_0}} \right)^n} } + {\omicron\left( {{{\left( {x – {x_0}} \right)}^n}} \right).}$

Since, by assumption, all of the first derivatives up to the $$\left( {n – 1} \right)$$th order are equal to zero, we obtain:

${f\left( x \right) – f\left( {{x_0}} \right) } = {\frac{{{f^{\left( n \right)}}\left( {{x_0}} \right)}}{{n!}}{\left( {x – {x_0}} \right)^n} } + {\omicron\left( {{{\left( {x – {x_0}} \right)}^n}} \right),}$

where the remainder term $$\omicron\left( {{{\left( {x – {x_0}} \right)}^n}} \right)$$ has a higher order of smallness than $$n.$$ As a result, the sign of the difference $$f\left( x \right) – f\left( {{x_0}} \right)$$ in the $$\delta$$-neighborhood of the point $${x_0}$$ will be determined by the sign of the $$n$$th term in the Taylor series:

${\text{sign}\left[ {f\left( x \right) – f\left( {{x_0}} \right)} \right] } = {\text{sign}\left[ {\frac{{{f^{\left( n \right)}}\left( {{x_0}} \right)}}{{n!}}{{\left( {x – {x_0}} \right)}^n}} \right]}$

or

${\text{sign} \left[ {f\left( x \right) – f\left( {{x_0}} \right)} \right] } = {\text{sign} \left[ {{f^{\left( n \right)}}\left( {{x_0}} \right){{\left( {x – {x_0}} \right)}^n}} \right].}$

If $$n$$ is an even number ($$n = 2k$$), then

${\forall \;x \in \left( {{x_0} – \delta ,{x_0} + \delta } \right) }\Rightarrow {{\left( {x – {x_0}} \right)^{2k}} > 0.}$

Consequently, in this case

${\text{sign}\left[ {f\left( x \right) – f\left( {{x_0}} \right)} \right] } = {\text{sign}{f^{\left( n \right)}}\left( {{x_0}} \right).}$

If $${f^{\left( n \right)}}\left( {{x_0}} \right) \gt 0$$ in the $$\delta$$-neighborhood of the point $${x_0}$$, then the following inequality holds:

${f\left( x \right) – f\left( {{x_0}} \right)} \gt 0.$

By definition, this means that $${x_0}$$ is a strict minimum point of the function $$f\left( x \right).$$

Similarly, if $${f^{\left( n \right)}}\left( {{x_0}} \right) \lt 0$$ in the $$\delta$$-neighborhood of the point $${x_0}$$, we have the inequality

${f\left( x \right) – f\left( {{x_0}} \right)} \lt 0,$

that corresponds to a strict maximum point.

If $$n$$ is an odd number $$\left(n = 2k + 1\right),$$ the degree of $${\left( {x – {x_0}} \right)^{2k + 1}}$$ will change sign when passing through the point $${x_0}.$$ Then it follows from the formula

${\text{sign} \left[ {f\left( x \right) – f\left( {{x_0}} \right)} \right] } = {\text{sign} \left[ {{f^{\left( n \right)}}\left( {{x_0}} \right){{\left( {x – {x_0}} \right)}^{2k + 1}}} \right]}$

that the difference $${f\left( x \right) – f\left( {{x_0}} \right)}$$ also changes sign when passing through $${x_0}.$$ In this case, the extremum at the point $${x_0}$$ does not exist.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Find the local extrema of the function $$f\left( x \right) = {e^{ – {x^2}}}.$$

### Example 2

Find the local (relative) extrema of the function $f\left( x \right) = – {x^2} + 4x – 3.$

### Example 3

Find the local extrema of the function $$f\left( x \right) = {x^3} – 12x.$$

### Example 4

Find the local extrema of the cubic function $f\left( x \right) = {x^3} – 3{x^2} – 9x + 2.$

### Example 5

Using the second derivative test, find the local extrema of the function $$f\left( x \right) = {x^3} – 9{x^2} + 24x – 7.$$

### Example 6

Find the local extrema of the function $$f\left( x \right) = {x^{\large\frac{1}{x}\normalsize}}.$$

### Example 7

Find the local extrema of the function $$f\left( x \right) = {x^2}{e^{ – x}}.$$

### Example 8

Find the local extrema points of the function $$f\left( x \right) = \left( {x – a} \right){e^x},$$ where $$a$$ is an arbitrary real number.

### Example 9

Find the local extrema of the function $$f\left( x \right) = x + \cos 2x.$$

### Example 10

Determine the local extrema of the function $$f\left( x \right) = \sin x – \cos x$$ on $$\left[ {0,2\pi } \right].$$

### Example 11

Determine the local extrema of the function $$f\left( x \right) = {x^2}\ln x.$$

### Example 12

Find the local extrema of the function $$f\left( x \right) = x\ln x.$$

### Example 13

Find the local extrema of the function $f\left( x \right) = {x^4} – 8{x^3} + 22{x^2} – 24x + 1.$

### Example 14

Find the local extrema of the function $f\left( x \right) = \frac{{{x^2}}}{{{x^4} + 16}}.$

### Example 15

Find the local extrema points of the function $$f\left( x \right) = \large{\frac{1}{{{x^2} – x}}}\normalsize.$$

### Example 16

Derive Snell’s law, which describes the refraction of light at the interface of two media.

### Example 17

Find the local extrema of the function $f\left( x \right) = x + \sqrt {1 – x} .$

### Example 18

Find the local extrema of the function $$f\left( x \right) = \sqrt {{x^2} + 1}.$$

### Example 19

Find the local extrema of $$f\left( x \right) = x\sqrt {1 – {x^2}} .$$

### Example 20

Find the local extrema of the function $f\left( x \right) = \frac{{{x^2} + x – 2}}{{x – 2}}.$

### Example 21

Find the local extrema of the function $f\left( x \right) = \frac{x}{2} – \arctan x.$

### Example 22

Determine the local extrema of the implicit function defined by the equation ${x^2} + {y^2} – 3xy + 5 = 0.$

### Example 23

Determine the local extrema of the function $f\left( x \right) = x + \frac{1}{x}.$

### Example 24

Find the local extrema of the function $$f\left( x \right) = {x^2} + \large{\frac{{16}}{{{x^2}}}}\normalsize.$$

### Example 25

Determine the local extrema of the function $$f\left( x \right) = x\,{\ln ^2}x.$$

### Example 26

Determine the minimum value of the function $$y = {x^{\sqrt x }}.$$

### Example 1.

Find the local extrema of the function $$f\left( x \right) = {e^{ – {x^2}}}.$$

Solution.

The function is defined and differentiable for all $$x \in \mathbb{R}.$$ We use the first derivative test. The derivative is given by

${f^\prime\left( x \right) = \left( {{e^{ – {x^2}}}} \right)^\prime }={ {e^{ – {x^2}}} \cdot \left( { – {x^2}} \right)^\prime }={ – 2x{e^{ – {x^2}}}.}$

Then we get

${f^\prime\left( x \right) = 0,}\;\; \Rightarrow {- 2x{e^{ – {x^2}}} = 0,}\;\; \Rightarrow {x = 0.}$

The derivative changes its sign as shown in the sign chart above.

Hence, the function has a maximum at $$x = 0.$$ The maximum value is

${{f_{\max }} = f\left( 0 \right) }={ {e^{ – {0^2}}} }={ {e^0} = 1.}$

### Example 2.

Find the local (relative) extrema of the function $f\left( x \right) = – {x^2} + 4x – 3.$

Solution.

This function is differentiable everywhere on the set $$\mathbb{R}.$$ Consequently, the extrema of the function are contained among its stationary points. Solve the equation $$f’\left( x \right) = 0:$$

${f’\left( x \right) = {\left( { – {x^2} + 4x – 3} \right)^\prime } } ={- 2x + 4,}$

${f’\left( x \right) = 0,\;\;}\Rightarrow {- 2x + 4 = 0,\;\;}\Rightarrow {x = 2.}$

The function has one stationary point $$x = 2.$$ Determine the sign of the derivative to the left and right of the point $$x = 2.$$ The derivative is positive for $$x \lt 2$$ and negative for $$x \gt 2$$. Thus, when passing through the point $$x = 2$$, the derivative changes sign from plus to minus. By the first derivative test, this means that $$x = 2$$ is a maximum point.

The maximum value (that is the value of the function at the maximum point) is equal to

${{f_{\max }} = f\left( 2 \right) } = { – {2^2} + 4 \cdot 2 – 3 = 1.}$

### Example 3.

Find the local extrema of the function $$f\left( x \right) = {x^3} – 12x.$$

Solution.

Determine the critical points. The first derivative is given by

${f^\prime\left( x \right) = \left( {{x^3} – 12x} \right)^\prime }={ 3{x^2} – 12.}$

It is equal to zero at the following points:

${f^\prime\left( x \right) = 0,}\;\; \Rightarrow {3{x^2} – 12 = 0,}\;\; \Rightarrow {{x^2} = 4,}\;\; \Rightarrow {{x_1} = – 2,{x_2} = 2.}$

These two points are critical since the function is defined and continuous over all $$x.$$ The derivative also exists for all $$x,$$ so there are no other critical points.

We use the Second Derivative Test:

${f^{\prime\prime}\left( x \right) = \left( {3{x^2} – 12} \right)^\prime }={ 6x,}$

${f^{\prime\prime}\left( { – 2} \right) = 6 \cdot \left( { – 2} \right) }={ -12 \lt 0,}$

${f^{\prime\prime}\left( {2} \right) = 6 \cdot 2 }={ 12 \gt 0.}$

Hence, $$x = -2$$ is a point of local maximum, and $$x = 2$$ is a point of local minimum.

Calculate the $$y-$$values for these points:

${{f_{\max }} = f\left( { – 2} \right) }={ {\left( { – 2} \right)^3} – 12 \cdot \left( { – 2} \right) }={ 16,}$

${{f_{\min }} = f\left( 2 \right) }={ {2^3} – 12 \cdot 2 }={ – 16.}$

${\text{local max:}\,\left( { -2, 16} \right);\,}\kern0pt {\text{local min:}\,\left( {2, -16} \right).}$

### Example 4.

Find the local extrema of the cubic function $f\left( x \right) = {x^3} – 3{x^2} – 9x + 2.$

Solution.

The function is differentiable on the whole set of real numbers. Therefore, the extremum points are contained among the stationary points (where the derivative is equal to zero).

We find these points:

${f’\left( x \right) = 0,\;\;}\Rightarrow {\left( {{x^3} – 3{x^2} – 9x + 2} \right) = 0,\;\;}\Rightarrow {3{x^2} – 6x – 9 = 0,\;\;}\Rightarrow {{x^2} – 2x – 3 = 0,\;\;}\Rightarrow {{x_1} = – 1,\;{x_2} = 2.}$

Substituting test values of $$x$$, we determine the sign of the derivative $$f’\left( x \right) = 3{x^2} – 6x – 9$$ in the corresponding intervals (Figure $$5$$).

As seen, when passing through the point $$x = – 1$$, the derivative changes sign from plus to minus. By the first derivative test, this point is a local maximum point. Similarly, we establish that $$x = 2$$ is a local minimum point.

We now determine the maximum and minimum values of the function:

${{f_{\max }} = f\left( { – 1} \right) } = {{\left( { – 1} \right)^3} – 3 \cdot {\left( { – 1} \right)^2} – 9 \cdot \left( { – 1} \right) + 2 = 7,}$

${{f_{\min }} = f\left( 2 \right) } = {{2^3} – 3 \cdot {2^2} – 9 \cdot 2 + 2 = – 20.}$

### Example 5.

Using the second derivative test, find the local extrema of the function $$f\left( x \right) = {x^3} – 9{x^2} + 24x – 7.$$

Solution.

The function is defined for all $$x.$$ Take the first derivative and determine the critical points:

${f^\prime\left( x \right) }={ \left( {{x^3} – 9{x^2} + 24x – 7} \right)^\prime }={ 3{x^2} – 18x + 24.}$

${f^\prime\left( x \right) = 0,}\;\; \Rightarrow {3{x^2} – 18x + 24 = 0,}\;\; \Rightarrow {3\left( {{x^2} – 6x + 8} \right) = 0,}\;\; \Rightarrow {3\left( {x – 2} \right)\left( {x – 4} \right) = 0,}\;\; \Rightarrow {{x_1} = 2,{x_2} = 4.}$

The second derivative is given by

${f^{\prime\prime}\left( x \right) = \left( {3{x^2} – 18x + 24} \right)^\prime }={ 6x – 18.}$

Determine the sign of the $$2$$nd derivative at the critical points:

${f^{\prime\prime}\left( {{x_1}} \right) }={ f^{\prime\prime}\left( 2 \right) }={ 6 \cdot 2 – 18 }={ – 6 \lt 0.}$

${f^{\prime\prime}\left( {{x_2}} \right) }={ f^{\prime\prime}\left( 4 \right) }={ 6 \cdot 4 – 18 }={ 6 \gt 0.}$

Hence, the point $${x_1} = 2$$ is a local maximum, and the point $${x_2} = 4$$ is a local minimum.

Compute the $$y-$$coordinates:

${f\left( {{x_1}} \right) }={ {2^3} – 9 \cdot {2^2} + 24 \cdot 2 – 7 }={ 13,}$

${f\left( {{x_2}} \right) }={ {4^3} – 9 \cdot {4^2} + 24 \cdot 4 – 7 }={ 9.}$

${\text{local max:}\,\left( {2,13} \right);\,\text{local min:}\,}\kern0pt{\left( {4,9} \right).}$

### Example 6.

Find the local extrema of the function $$f\left( x \right) = {x^{\large\frac{1}{x}\normalsize}}.$$

Solution.

This function belongs to the family of power-exponential functions. In general, they have the form of $$y = g{\left( x \right)^{h\left( x \right)}}.$$ It is usually assumed that the domain of power-exponential functions satisfies the condition $$g\left( x \right) \gt 0.$$ (In some special cases, the base $$g\left( x \right)$$ can be negative − for example, if $$h = {\large\frac{1}{3}\normalsize}.$$) In our case we suppose that $$x \gt 0.$$ It follows from here that the function takes only positive values.

Determine the derivative:

${f’\left( x \right) = {\left( {{x^{\large\frac{1}{x}\normalsize}}} \right)^\prime } } = {{\left( {{e^{\ln {x^{\frac{1}{x}}}}}} \right)^\prime } } = {{e^{\ln {x^{\frac{1}{x}}}}} \cdot {\left( {\ln {x^{\large\frac{1}{x}\normalsize}}} \right)^\prime } } = {{x^{\large\frac{1}{x}\normalsize}} \cdot {\left( {\frac{1}{x}\ln x} \right)^\prime } } = {{x^{\large\frac{1}{x}\normalsize}} \cdot\kern0pt {\left[ {{{\left( {\frac{1}{x}} \right)}^\prime }\ln x + \frac{1}{x}{{\left( {\ln x} \right)}^\prime }} \right]} } = {{x^{\large\frac{1}{x}\normalsize}} \cdot\kern0pt {\left[ {\left( { – \frac{1}{{{x^2}}}} \right) \cdot \ln x + \frac{1}{x} \cdot \frac{1}{x}} \right]} } = {\frac{{{x^{\large\frac{1}{x}\normalsize}}}}{{{x^2}}}\left( {1 – \ln x} \right) } = {{x^{\large\frac{1}{x}\normalsize – 2}}\left( {1 – \ln x} \right).}$

Note: The derivative of the function $$f\left( x \right) = {x^{\large\frac{1}{x}\normalsize}}$$ can also be found using logarithmic differentiation.

Calculate the critical points:

${f’\left( x \right) = 0,\;\;}\Rightarrow {{x^{\large\frac{1}{x}\normalsize – 2}}\left( {1 – \ln x} \right) = 0,\;\;}\Rightarrow {1 – \ln x = 0,\;\;}\Rightarrow {\ln x = 1,\;\;}\Rightarrow {x = e.}$

In the left neighborhood of $$x = e,$$ the derivative is positive, and in the right neighborhood it is negative. Consequently, the function attains a local maximum at the point $$x = e.$$ The maximum value is equal

${{f_{\max }} = f\left( {x = e} \right) } = {{e^{\large\frac{1}{e}\normalsize}} \approx 1,445.}$

### Example 7.

Find the local extrema of the function $$f\left( x \right) = {x^2}{e^{ – x}}.$$

Solution.

The function is defined and differentiable on the whole set $$\mathbb{R}.$$ Calculate its derivative:

${f’\left( x \right) = {\left( {{x^2}{e^{ – x}}} \right)^\prime } } = {{\left( {{x^2}} \right)^\prime }{e^{ – x}} + {x^2}{\left( {{e^{ – x}}} \right)^\prime } } = {2x{e^{ – x}} – {x^2}{e^{ – x}} = x{e^{ – x}}\left( {2 – x} \right).}$

Find the roots of the equation $$f’\left( x \right) = 0:$$

${x{e^{ – x}}\left( {2 – x} \right) = 0,\;\;}\Rightarrow {{x_1} – 0,\;{x_2} = 2.}$

When passing through these points, the derivative changes sign as shown above in Figure $$6.$$

Hence, at the point $$x = 0,$$ the function has a minimum, and at the point $$x = 0$$ it has a maximum. The minimum and maximum values, respectively, are equal to:
${{f_{\min }} = f\left( 0 \right) } = {{0^2}{e^{ – 0}} = 0,}$
${{f_{\max }} = f\left( 2 \right) } = {{2^2}{e^{ – 2}} } = {\frac{4}{{{e^2}}} \approx 0,541.}$