# Calculus

## Applications of the Derivative # Local Extrema of Functions

## Solved Problems

### Example 15.

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

Solution.

The domain of the function is given by

${x^2} - x \ne 0,\;\; \Rightarrow x\left( {x - 1} \right) \ne 0,\;\; \Rightarrow x \ne 0,1.$

Find the derivative:

$f^\prime\left( x \right) = \left( {\frac{1}{{{x^2} - x}}} \right)^\prime = \left( {{{\left( {{x^2} - x} \right)}^{ - 1}}} \right)^\prime = - {\left( {{x^2} - x} \right)^{ - 2}} \cdot \left( {2x - 1} \right) = - \frac{{2x - 1}}{{{{\left( {{x^2} - x} \right)}^2}}}.$

We have one critical point:

$f^\prime\left( x \right) = 0,\;\; \Rightarrow - \frac{{2x - 1}}{{{{\left( {{x^2} - x} \right)}^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x - 1 = 0}\\ {{{\left( {{x^2} - x} \right)}^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = \frac{1}{2}}\\ {x \ne 0,1} \end{array}} \right.,\;\; \Rightarrow x = \frac{1}{2}.$

Draw a sign chart for the first derivative (see above). We see from the chart that $$x = \frac{1}{2}$$ is a local maximum point. Its $$y-$$value is

$f\left( {\frac{1}{2}} \right) = \frac{1}{{{{\left( {\frac{1}{2}} \right)}^2} - \frac{1}{2}}} = \frac{1}{{\frac{1}{4} - \frac{1}{2}}} = \frac{1}{{\left( { - \frac{1}{4}} \right)}} = - 4.$

$\text{local max:}\,\left( {\frac{1}{2}, -4} \right).$

### Example 16.

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

Solution.

The law of refraction of light was established in $$1621$$ by the Dutch physicist and mathematician Willebrord Snell (Snellius) $$\left(1580-1626\right)$$ and independently later by the French scientist Rene Descartes $$\left(1596-1650\right).$$

In $$1657$$, The French mathematician Pierre de Fermat $$\left({1601-1665}\right)$$ then derived this law using the principle of least action. It was one of the first problems of calculus of variations.

At that time the derivatives were not yet known, and Fermat used for proving his own method of analysis of infinitesimals (adequality). Now this problem is easily solved with the help of the derivative.

According to Fermat's principle of least action, a ray of light, which travels between any two points follows the least time path of propagation.

Consider two media with a flat interface between them (Figure $$14$$). Let the light travel from point $$A$$ to point $$B.$$

Suppose that the angle of incidence (the angle between the incident ray and the normal to the interface) in the first medium is $${\alpha _1}$$ and the angle of refraction in the second medium is $${\alpha _2}.$$ The speed of light in the first and second medium are, respectively, $${v_1}$$ and $${v_1}.$$

Let the ray intersect the boundary between the media at the point with the coordinate $$x.$$ The task is to determine the value of $$x,$$ at which the light propagation time is minimum.

If the $$y$$-coordinate of the point $$A$$ is $${y_1},$$ then the signal propagation time in the first medium is equal to $${t_1} = {\frac{{\sqrt {y_1^2 + {x^2}} }}{{{v_1}}}}.$$ Similarly, the signal propagation time in the second medium is given by the formula $${t_2} = {\frac{{\sqrt {y_2^2 + {{\left( {L - x} \right)}^2}} }}{{{v_2}}}},$$ where $$L$$ denotes the $$x$$-coordinate of the point $$B.$$ The total time of propagation is a function of $$x:$$

$t\left( x \right) = {t_1} + {t_2} = \frac{{\sqrt {y_1^2 + {x^2}} }}{{{v_1}}} + \frac{{\sqrt {y_2^2 + {{\left( {L - x} \right)}^2}} }}{{{v_2}}}.$

Determine the extremum of this function. The derivative has the form:

$t'\left( x \right) = \left[ {\frac{{\sqrt {y_1^2 + {x^2}} }}{{{v_1}}} + \frac{{\sqrt {y_2^2 + {{\left( {L - x} \right)}^2}} }}{{{v_2}}}} \right]^\prime = \frac{x}{{{v_1}\sqrt {y_1^2 + {x^2}} }} - \frac{{L - x}}{{{v_2}\sqrt {y_2^2 + {{\left( {L - x} \right)}^2}} }}.$

Since

$\frac{x}{{\sqrt {y_1^2 + {x^2}} }} = \sin {\alpha _1},\;\;\frac{{L - x}}{{\sqrt {y_2^2 + {{\left( {L - x} \right)}^2}} }} = \sin {\alpha _2},$

then there exists an extremum at the value $$x = {x_0},$$ when

$\frac{{\sin {\alpha _1}}}{{{v_1}}} - \frac{{\sin {\alpha _2}}}{{{v_2}}} = 0\;\;\text{or}\;\;\frac{{\sin {\alpha _1}}}{{{v_1}}} = \frac{{\sin {\alpha _2}}}{{{v_2}}}.$

The derivative $$t'\left( x \right)$$ in the left neighborhood of the critical point $${x_0}$$ is negative, and to the right is positive. Hence, the extremum point is a minimum point. Thus, we got the law of refraction, which is written as

$\frac{{\sin {\alpha _1}}}{{\sin {\alpha _2}}} = \frac{{{v_1}}}{{{v_2}}}.$

### Example 17.

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

Solution.

The function is defined and differentiable for $$x \le 1.$$ Find the derivative:

$f'\left( x \right) = {\left( {x + \sqrt {1 - x} } \right)^\prime } = 1 + \frac{1}{{2\sqrt {1 - x} }} \cdot \left( { - 1} \right) = \frac{{2\sqrt {1 - x} - 1}}{{2\sqrt {1 - x} }}.$

Determine the critical points:

$f'\left( x \right) = 0,\;\; \Rightarrow \frac{{2\sqrt {1 - x} - 1}}{{2\sqrt {1 - x} }} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2\sqrt {1 - x} - 1 = 0}\\ {2\sqrt {1 - x} \ne 0} \end{array}} \right.,\;\; \Rightarrow \sqrt {1 - x} = \frac{1}{2},\;\; \Rightarrow 1 - x = \frac{1}{4},\;\; \Rightarrow x = \frac{3}{4}.$

Thus, we have one stationary point $$x = {\frac{3}{4}}$$ and one critical point $$x = 1$$ (in which the derivative does not exist).

Investigate the stationary point $$x = {\frac{3}{4}}.$$ In the left neighborhood of this point the derivative is positive, and in the right neighborhood it is negative, i.e. we deal here with a local maximum. Its value is equal to

${f_{\max }} = f\left( {\frac{3}{4}} \right) = \frac{3}{4} + \sqrt {1 - \frac{3}{4}} = \frac{5}{4}.$

As to the critical point $$x = 1,$$ then it exists on the boundary of the domain, so it cannot be a local extremum.

### Example 18.

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

Solution.

We use the $$1$$st Derivative Test to find the extrema point.

Note that the function is defined for all $$x \in \mathbb{R}.$$ Take the derivative and equate it to zero:

$f^\prime\left( x \right) = \left( {\sqrt {{x^2} + 1} } \right)^\prime = \frac{1}{{2\sqrt {{x^2} + 1} }} \cdot \left( {{x^2} + 1} \right)^\prime = \frac{{\cancel{2}x}}{{\cancel{2}\sqrt {{x^2} + 1} }} = \frac{x}{{\sqrt {{x^2} + 1} }}.$
$f^\prime\left( x \right) = 0, \;\; \Rightarrow \frac{x}{{\sqrt {{x^2} + 1} }} = 0,\;\; \Rightarrow x = 0.$

Now we plot this critical point on a number line and determine the intervals where the function is increasing or decreasing.

We see from the sign chart that $$f\left( x \right)$$ has a local minimum at $$x = 0.$$ The $$y-$$coordinate of the local minimum is

$f\left( 0 \right) = \sqrt {{0^2} + 1} = 1.$

### Example 19.

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

Solution.

First we determine the domain of the function:

$1 - {x^2} \ge 0,\;\; \Rightarrow {x^2} \le 1,\;\; \Rightarrow x \in \left[ { - 1,1} \right].$

Find the derivative of the function using the product rule and the chain rule:

$f^\prime\left( x \right) = \left( {x\sqrt {1 - {x^2}} } \right)^\prime = x^\prime \cdot \sqrt {1 - {x^2}} + x \cdot \left( {\sqrt {1 - {x^2}} } \right)^\prime = 1 \cdot \sqrt {1 - {x^2}} + x \cdot \frac{{\left( { - 2x} \right)}}{{2\sqrt {1 - {x^2}} }} = \sqrt {1 - {x^2}} - \frac{{{x^2}}}{{\sqrt {1 - {x^2}} }} = \frac{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2} - {x^2}}}{{\sqrt {1 - {x^2}} }} = \frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }}.$

Calculate the critical points:

$f^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{1 - 2{x^2}}}{{\sqrt {1 - {x^2}} }} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {1 - 2{x^2} = 0}\\ {\sqrt {1 - {x^2}} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} = \frac{1}{2}}\\ {x \ne \pm 1} \end{array}} \right.,\;\; \Rightarrow {x_{1,2}} = \pm \frac{1}{{\sqrt 2 }} = \pm \frac{{\sqrt 2 }}{2}.$

So we have two critical points $${x_1} = - \frac{{\sqrt 2 }}{2}$$ and $${x_2} = \frac{{\sqrt 2 }}{2}.$$ Both these points belong to the domain of the function.

Draw a sign chart for $$f^\prime\left( x \right)$$ (see above). We see that the first point $${x_1} = - \frac{{\sqrt 2 }}{2}$$ is a point of local minimum and $${x_2} = \frac{{\sqrt 2 }}{2}$$ is a point of local maximum.

Compute the $$y-$$coordinates of the extrema points:

$f\left( { - \frac{{\sqrt 2 }}{2}} \right) = \left( { - \frac{{\sqrt 2 }}{2}} \right) \cdot \sqrt {1 - {{\left( { - \frac{{\sqrt 2 }}{2}} \right)}^2}} = \left( { - \frac{{\sqrt 2 }}{2}} \right) \cdot \sqrt {1 - \frac{1}{2}} = \left( { - \frac{{\sqrt 2 }}{2}} \right) \cdot \frac{{\sqrt 2 }}{2} = - \frac{1}{2}.$
$f\left( {\frac{{\sqrt 2 }}{2}} \right) = \frac{{\sqrt 2 }}{2} \cdot \sqrt {1 - {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}} = \frac{{\sqrt 2 }}{2} \cdot \sqrt {1 - \frac{1}{2}} = \frac{{\sqrt 2 }}{2} \cdot \frac{{\sqrt 2 }}{2} = \frac{1}{2}.$

$\text{local min:}\,\left( { - \frac{{\sqrt 2 }}{2}, - \frac{1}{2}} \right);\;\text{local max :}\,\left( { \frac{{\sqrt 2 }}{2}, \frac{1}{2}} \right).$

### Example 20.

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

Solution.

Calculate the derivative and find the critical points of the function:

$f'\left( x \right) = \left( {\frac{{{x^2} + x - 2}}{{x - 2}}} \right)^\prime = \frac{{\left( {2x + 1} \right)\left( {x - 2} \right) - \left( {{x^2} + x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{x\left( {x - 4} \right)}}{{{{\left( {x - 2} \right)}^2}}}.$

The function has two stationary points $$x = 0,$$ $$x = 4$$ (where the derivative is zero) and a critical point $$x = 2$$ (in which the derivative does not exist).

Next we investigate how the derivative changes sign when passing through these points (Figure $$17$$).

As it can be seen, there is a local maximum at the point $$x = 0.$$ Its value is

${f_{\max }} = f\left( 0 \right) = \frac{{{0^2} + 0 - 2}}{{0 - 2}} = 1.$

At the point $$x = 4$$, the function has a minimum equal

${f_{\min }} = f\left( 4 \right) = \frac{{{4^2} + 4 - 2}}{{4 - 2}} = 9.$

At the point $$x = 2$$, the function has a discontinuity, so there is no extremum here.

A schematic view of this function is shown in Figure $$18.$$

### Example 21.

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

Solution.

The function is defined and differentiable on the whole set $$\mathbb{R}.$$ Find its stationary points:

$f'\left( x \right) = {\left( {\frac{x}{2} - \arctan x} \right)^\prime } = \frac{1}{2} - \frac{1}{{1 + {x^2}}} = \frac{{1 + {x^2} - 2}}{{2\left( {1 + {x^2}} \right)}} = \frac{{{x^2} - 1}}{{2\left( {1 + {x^2}} \right)}};$
$f'\left( x \right) = 0,\;\; \Rightarrow \frac{{{x^2} - 1}}{{2\left( {1 + {x^2}} \right)}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} - 1 = 0}\\ {2\left( {1 + {x^2}} \right) \ne 0} \end{array}} \right.,\;\; \Rightarrow {x_{1,2}} = \pm 1.$

To determine the extremum type at these points, we use the second derivative test. Calculate the second derivative:

$f^{\prime\prime}\left( x \right) = {\left( {\frac{{{x^2} - 1}}{{2 + 2{x^2}}}} \right)^\prime } = \frac{{2x\left( {2 + 2{x^2}} \right) - 4x\left( {{x^2} - 1} \right)}}{{{{\left( {2 + 2{x^2}} \right)}^2}}} = \frac{{\color{red}{4x} + \cancel{\color{blue}{4{x^3}}} - \cancel{\color{blue}{4{x^3}}} + \color{red}{4x}}}{{4{{\left( {1 + {x^2}} \right)}^2}}} = \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}.$

Substituting the stationary points $$x = -1,$$ $$x = 1,$$ we find:

$f^{\prime\prime}\left( { - 1} \right) = - \frac{1}{2} \lt 0,\;\;f^{\prime\prime}\left( 1 \right) = \frac{1}{2} \gt 0.$

Thus, the point $$x = -1$$ is a local maximum point, and the point $$x = 1$$ is a local minimum point. The respective values of the function at these points are

${f_{\max }} = f\left( { - 1} \right) = - \frac{1}{2} - \arctan \left( { - 1} \right) = - \frac{1}{2} + \arctan 1 = \frac{\pi }{4} - \frac{1}{2} \approx 0,29;$
${f_{\min }} = f\left( 1 \right) = \frac{1}{2} - \arctan 1 = - \left( {\frac{\pi }{4} - \frac{1}{2}} \right) \approx - 0,29.$

### Example 22.

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

Solution.

We begin our analysis by calculating the derivative of the implicit function. Differentiating both sides of the equation with respect to $$x,$$ we obtain:

${\left( {{x^2} + {y^2} - 3xy + 5} \right)_x}^\prime = 0,\;\; \Rightarrow 2x + 2yy' - 3y - 3xy' = 0,\;\; \Rightarrow 2x - 3y + y'\left( {2y - 3x} \right) = 0,\;\; \Rightarrow y' = - \frac{{2x - 3y}}{{2y - 3x}} = \frac{{2x - 3y}}{{3x - 2y}}.$

As it can be seen, the derivative becomes infinite when $$3x - 2y = 0,$$ i.e. $$y = {\frac{3x}{2}}.$$ This occurs at the following critical points:

${x^2} + {\left( {\frac{{3x}}{2}} \right)^2} - 3x \cdot \frac{{3x}}{2} + 5 = 0,\;\; \Rightarrow {x^2} + \frac{{9{x^2}}}{4} - \frac{{9{x^2}}}{2} + 5 = 0,\;\; \Rightarrow - \frac{{5{x^2}}}{4} + 5 = 0,\;\; \Rightarrow {x^2} = 4,\;\; \Rightarrow x = \pm 2.$

Examine the domain of the function $$y\left( x \right).$$ The given equation can be rewritten in the form

${y^2} - 3xy + {x^2} + 5 = 0.$

To express $$y\left( x \right)$$ explicitly, it is necessary to solve the quadratic equation. Its discriminant is given by

$D = {\left( { - 3x} \right)^2} - 4\left( {{x^2} + 5} \right) = 9{x^2} - 4{x^2} - 20 = 5{x^2} - 20 = 5\left( {{x^2} - 4} \right).$

It is clear that the function $$y\left( x \right)$$ exists if $${x^2} \ge 4.$$ It follows from here that the critical points $$x = \pm 2$$ are on the boundary of the domain of the function $$y\left( x \right)$$ and, by definition, they cannot be local extrema.

Now we find the stationary points within the domain of the function (where the derivative $$y'\left( x \right)$$ is equal to zero):

$y' = \frac{{2x - 3y}}{{3x - 2y}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2x - 3y = 0}\\ {3x - 2y \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {y = \frac{2}{3}x}\\ {y \ne \frac{3}{2}x} \end{array}} \right..$

Substituting the expression for $$y$$ in the original equation, we get:

${x^2} + {\left( {\frac{2}{3}x} \right)^2} - 3x \cdot \frac{2}{3}x + 5 = 0,\;\; \Rightarrow {x^2} + \frac{{4{x^2}}}{9} - 2{x^2} + 5 = 0,\;\; \Rightarrow - \frac{{5{x^2}}}{9} + 5 = 0,\;\; \Rightarrow {x^2} = 9,\;\; \Rightarrow {x_{1,2}} = \pm 3.$

So we have two stationary points:

${x_1} = -3,\;\;y\left( { - 3} \right) = - 2;$
${x_2} = 3,\;\;y\left( {3} \right) = 2.$

To figure out the type of extremum at these points, we use the second derivative test. Calculate the second derivative:

$y^{\prime\prime} = {\left( {\frac{{2x - 3y}}{{3x - 2y}}} \right)^\prime } = \frac{{{5y} - {5xy'}}}{{{{\left( {3x - 2y} \right)}^2}}}.$

Substituting the coordinates of the stationary points, we find:

$y^{\prime\prime}\left( { - 3, - 2} \right) = \frac{{5 \cdot \left( { - 2} \right) - 5 \cdot \left( { - 3} \right) \cdot 0}}{{{{\left( {3 \cdot \left( { - 3} \right) - 2 \cdot \left( { - 2} \right)} \right)}^2}}} = - \frac{2}{5} \lt 0,$
$y^{\prime\prime}\left( {3,2} \right) = \frac{{5 \cdot 2 - 5 \cdot 3 \cdot 0}}{{{{\left( {3 \cdot 3 - 2 \cdot 2} \right)}^2}}} = \frac{2}{5} \gt 0.$

Hence, the point $$\left( { - 3, - 2} \right)$$ is a local maximum of the implicit function, and the point $$\left( { 3, 2} \right)$$ is a local minimum. This function is a hyperbola. Its schematic view is given in Figure $$19.$$

### Example 23.

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

Solution.

The function is defined and differentiable for all real $$x \ne 0.$$ Find the derivative and calculate the stationary points:

$f'\left( x \right) = \left( {x + \frac{1}{x}} \right)^\prime = 1 - \frac{1}{{{x^2}}} = \frac{{{x^2} - 1}}{{{x^2}}};$
$f'\left( x \right) = 0,\;\; \Rightarrow \frac{{{x^2} - 1}}{{{x^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} - 1 = 0}\\ {{x^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow {x_{1,2}} = \pm 1.$

Find the second derivative:

$f^{\prime\prime}\left( x \right) = \left( {1 - \frac{1}{{{x^2}}}} \right)^\prime = \left( { - {x^{ - 2}}} \right)^\prime = 2{x^{ - 3}} = \frac{2}{{{x^3}}}.$

Substituting $$x = \pm 1,$$ we determine the sign of the second derivative at these points:

$f^{\prime\prime}\left( { - 1} \right) = \frac{2}{{{{\left( { - 1} \right)}^3}}} = - 2 \lt 0,\;\;f^{\prime\prime}\left( 1 \right) = \frac{2}{{{1^3}}} = 2 \gt 0.$

According to the second derivative test, we obtain that $$x = -1$$ is a local maximum point, and $$x = 1$$ is a local minimum point. The values of the function at the given points are equal

${f_{\max }} = f\left( { - 1} \right) = - 1 + \frac{1}{{\left( { - 1} \right)}} = - 2,\;\;{f_{\min }} = f\left( 1 \right) = 1 + \frac{1}{1} = 2.$

The graph of the function is schematically shown in Figure $$20.$$

### Example 24.

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

Solution.

The function is defined for all $$x \in \mathbb{R},$$ except the point $$x = 0$$ where it has a discontinuity.

Take the derivative:

$f^\prime\left( x \right) = \left( {{x^2} + \frac{{16}}{{{x^2}}}} \right)^\prime = \left( {{x^2} + 16{x^{ - 2}}} \right)^\prime = 2x - 32{x^{ - 3}} = 2x - \frac{{32}}{{{x^3}}} = \frac{{2{x^4} - 32}}{{{x^3}}}.$

Determine the critical points:

$f^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{2{x^4} - 32}}{{{x^3}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2{x^4} - 32 = 0}\\ {{x^3} \ne 0} \end{array}} \right.,\;\; \Rightarrow {x_{1,2}} = \pm 2.$

Draw a sign chart for $$f^\prime\left( x \right)$$ (see the figure above).

According to the $$1$$st Derivative Test, the function has local minima at two points: $${x_1} = - 2$$ and $${x_1} = 2.$$ Calculate the $$y-$$values at these points:

$f\left( { - 2} \right) = {\left( { - 2} \right)^2} + \frac{{16}}{{{{\left( { - 2} \right)}^2}}} = 8.$

As the function is even, we can immediately write:

$f\left( 2 \right) = f\left( { - 2} \right) = 8.$

$\text{local min:}\,\left( { - 2,8} \right),\,\left( {2,8} \right).$

### Example 25.

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

Solution.

This function is defined and differentiable for $$x \gt 0.$$ Find the stationary points of the function:

$f'\left( x \right) = \left( {x{{\ln }^2}x} \right)^\prime = {\left( x \right)^\prime }{\ln ^2}x + x{\left( {{{\ln }^2}x} \right)^\prime } = {\ln ^2}x + x \cdot 2\ln x \cdot \frac{1}{x} = {\ln ^2}x + 2\ln x = \ln x\left( {\ln x + 2} \right);$
$f'\left( x \right) = 0,\;\; \Rightarrow \ln x\left( {\ln x + 2} \right) = 0.$

The equation has two solutions:

$1)\;\ln x = 0,\;\; \Rightarrow {x_1} = 1;$
$2)\;\ln x + 2 = 0,\;\;\Rightarrow \ln x = - 2,\;\;\Rightarrow {x_2} = {e^{ - 2}} = \frac{1}{{{e^2}}} \approx 0,135.$

To figure out the type of extremum of the stationary points, it is convenient to use the second derivative test. Calculate the second derivative:

$f^{\prime\prime}\left( x \right) = {\left( {{{\ln }^2}x + 2\ln x} \right)^\prime } = 2\ln x \cdot \frac{1}{x} + \frac{2}{x} = \frac{2}{x}\left( {\ln x + 1} \right).$

At $$x = 1$$ and $$x = {\frac{1}{{{e^2}}}},$$ the second derivative has the following values:

$f^{\prime\prime}\left( 1 \right) = \frac{2}{1}\left( {\ln 1 + 1} \right) = 2 \gt 0,$
$f^{\prime\prime}\left( {\frac{1}{{{e^2}}}} \right) = \frac{2}{{\frac{1}{{{e^2}}}}}\left( {\ln \frac{1}{{{e^2}}} + 1} \right) = 2{e^2}\left( { - 2 + 1} \right) = -2{e^2} \lt 0.$

Thus, the function has a local minimum at$$x = 1$$ and a local maximum at $$x = {\frac{1}{{{e^2}}}}.$$ Compute the minimum and maximum values of the function:

${f_{\min }} = f\left( 1 \right) = 1 \cdot {\ln ^2}1 = 0,$
${f_{\max }} = f\left( {\frac{1}{{{e^2}}}} \right) = \frac{1}{{{e^2}}}{\ln ^2}\left( {\frac{1}{{{e^2}}}} \right) = \frac{4}{{{e^2}}} \approx 0,541.$

This function is shown schematically in Figure $$22.$$

### Example 26.

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

Solution.

This function is defined at $$x \gt 0.$$ We take the derivative using logarithmic differentiation:

$y = {x^{\sqrt x }},\;\; \Rightarrow \ln y = \ln \left( {{x^{\sqrt x }}} \right),\;\; \Rightarrow \ln y = \sqrt x \ln x,\;\; \Rightarrow \left( {\ln y} \right)^\prime = \left( {\sqrt x \ln x} \right)^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = \left( {\sqrt x } \right)^\prime \cdot \ln x + \sqrt x \cdot \left( {\ln x} \right)^\prime,\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{1}{{2\sqrt x }} \cdot \ln x + \sqrt x \cdot \frac{1}{x},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{\ln x}}{{2\sqrt x }} + \frac{1}{{\sqrt x }},\;\; \Rightarrow \frac{{y^\prime}}{y} = \frac{{\ln x + 2}}{{2\sqrt x }},\;\; \Rightarrow y^\prime = \frac{{\ln x + 2}}{{2\sqrt x }}{x^{\sqrt x }}.$

Find the critical points on $$\left( {0, + \infty } \right):$$

$y^\prime = 0,\;\; \Rightarrow \frac{{\ln x + 2}}{{2\sqrt x }}{x^{\sqrt x }} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\ln x + 2 = 0}\\ {x \ne 0} \end{array}} \right.,\;\; \Rightarrow \ln x = - 2,\;\; \Rightarrow x = {e^{ - 2}} = \frac{1}{{{e^2}}}.$

As it follows from the sign chart (see above), the function has a local minimum point at $$x = \frac{1}{{{e^2}}}.$$

Calculate the $$y-$$value of the local minimum:

${y_{\min }} = y\left( {\frac{1}{{{e^2}}}} \right) = {\left( {\frac{1}{{{e^2}}}} \right)^{\sqrt {\frac{1}{{{e^2}}}} }} = {\left( {\frac{1}{{{e^2}}}} \right)^{\frac{1}{e}}} = {\left( {{e^{ - 2}}} \right)^{\frac{1}{e}}} = {e^{ - \frac{2}{e}}} = \frac{1}{{{e^{\frac{2}{e}}}}} \approx 0.48$