# Calculus

## Applications of the Derivative # Curve Sketching

## Solved Problems

Sketch graphs of the following functions (Examples 5−14).

### Example 5.

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

Solution.

The function is defined for all $$x$$ except the point $$x = 1$$ where it has a discontinuity.

Find the $$y-$$intercept:

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

The function is negative for $$x \lt 1$$ and positive for $$x \gt 1,$$ but it has no $$x-$$intercepts.

Look for vertical asymptote near $$x = 1:$$

$\lim\limits_{x \to 1 - 0} f\left( x \right) = \lim\limits_{x \to 1 - 0} \frac{{{x^2} + 1}}{{x - 1}} = - \infty ;$
$\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \frac{{{x^2} + 1}}{{x - 1}} = \infty .$

There is a vertical asymptote at $$x = 1.$$

Rewrite the function in the form

$f\left( x \right) = \frac{{{x^2} + 1}}{{x - 1}} = \frac{{{x^2} - x + x - 1 + 2}}{{x - 1}} = \frac{{x\left( {x - 1} \right) + x - 1 + 2}}{{x - 1}} = x + 1 + \frac{2}{{x - 1}},$

where $$\frac{2}{{x - 1}} \to 0$$ as $$x \to \infty.$$ Hence the function has an oblique asymptote $$y = x + 1.$$

Take the first derivative:

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

Determine the critical points:

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

${x^2} - 2x - 1 = 0,\;\; \Rightarrow D = {\left( { - 2} \right)^2} - 4 \cdot \left( { - 1} \right) = 8,;\; \Rightarrow {x_{1,2}} = \frac{{2 \pm \sqrt 8 }}{2} = 1 \pm \sqrt 2 .$

Thus, the function has two critical points: $${x_1} = 1 - \sqrt 2 \approx - 0.41$$ and $${x_2} = 1 + \sqrt 2 \approx 2.41$$

The second derivative is written as

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

We see that the function is concave downward at $$x \lt 1$$ and concave upward at $$x \gt 1,$$ though it has no inflection points.

Draw a sign chart for the function and its derivatives (Figure $$5a$$).

The point $$x = 1 - \sqrt 2$$ is a local maximum, and the point $$x = 1 + \sqrt 2$$ is a local minimum. Calculate their $$y-$$coordinates:

$\require{cancel} f\left( {1 - \sqrt 2 } \right) = \frac{{{{\left( {1 - \sqrt 2 } \right)}^2} + 1}}{{\cancel{1} - \sqrt 2 - \cancel{1}}} = \frac{{1 - 2\sqrt 2 + 2 + 1}}{{ - \sqrt 2 }} = \frac{{4 - 2\sqrt 2 }}{{ - \sqrt 2 }} = \frac{{4 - 4\sqrt 2 }}{2} = 2\left( {1 - \sqrt 2 } \right) \approx {- 0.83}$
$f\left( {1 + \sqrt 2 } \right) = \frac{{{{\left( {1 + \sqrt 2 } \right)}^2} + 1}}{{\cancel{1} + \sqrt 2 - \cancel{1}}} = \frac{{1 + 2\sqrt 2 + 2 + 1}}{{ \sqrt 2 }} = \frac{{4 + 2\sqrt 2 }}{{\sqrt 2 }} = \frac{{4 + 4\sqrt 2 }}{2} = 2\left( {1 + \sqrt 2 } \right) \approx {4.83}$

Now we can sketch a graph of the function (Figure $$5b$$).

### Example 6.

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

Solution.

The function is defined for all $$x$$ except the points $$x = \pm 1$$ where it has discontinuities.

The function has no roots $$\left(x-\right.$$intercepts. Calculate the $$y-$$intercept:

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

We see that the function is positive for $$x \in \left( { - 1,1} \right)$$ and negative for $$\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right).$$

The function is even since

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

Determine the vertical asymptotes:

$\lim\limits_{x \to - 1 - 0} f\left( x \right) = \lim\limits_{x \to - 1 - 0} \frac{1}{{1 - {x^2}}} = - \infty ;$
$\lim\limits_{x \to - 1 + 0} f\left( x \right) = \lim\limits_{x \to - 1 + 0} \frac{1}{{1 - {x^2}}} = + \infty .$

As the function is even, we have

$\lim\limits_{x \to 1 - 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \frac{1}{{1 - {x^2}}} = + \infty ;$
$\lim\limits_{x \to 1 + 0} f\left( x \right) = \lim\limits_{x \to 1 + 0} \frac{1}{{1 - {x^2}}} = - \infty .$

Hence, there are two vertical asymptotes $$x = -1$$ and $$x = 1.$$

Look for horizontal asymptotes:

$\lim\limits_{x \to \pm \infty } f\left( x \right) = \lim\limits_{x \to \pm \infty } \frac{1}{{1 - {x^2}}} = 0.$

So, $$y = 0$$ (the $$x-$$axis) is a horizontal asymptote.

Since the function has a two-sided horizontal asymptote, it does not have any oblique asymptote.

Take the derivative:

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

Find the critical points:

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

Note that the points $$x = \pm 1$$ are not critical as the function is undefined there.

The point $$x = 0$$ is a local minimum. Its $$y-$$value is

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

Hence, there is a local minimum at $$\left( {0,1} \right).$$

Consider now the $$2$$nd derivative, concavity and inflection points.

$f^{\prime\prime}\left( x \right) = \left( {\frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^2}}}} \right)^\prime = \frac{{2 + 6{x^2}}}{{{{\left( {1 - {x^2}} \right)}^3}}}.$

Solve the equation

$f^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{2 + 6{x^2}}}{{{{\left( {1 - {x^2}} \right)}^3}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {2 + 6{x^2} = 0}\\ {{{\left( {1 - {x^2}} \right)}^3} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} = - \frac{1}{3}}\\ {{x^2} \ne \pm 1} \end{array}} \right.,\;\; \Rightarrow x \in \varnothing.$

We see that $$f^{\prime\prime}$$ is nowhere zero, but it does not exist at $$x = \pm 1.$$ Since these points do not belong to the domain, we conclude that the function has no inflection points.

The graph of the given function is sketched in Figure $$6b$$.

### Example 7.

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

Solution.

The function is defined for all $$x$$ except the point $$x = 0$$ where it has a discontinuity.

Find the $$x-$$intercepts:

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

The function is neither even, nor odd.

Check for vertical asymptotes:

$\lim\limits_{x \to 0 - 0} f\left( x \right) = \lim\limits_{x \to 0 - 0} \left( {{x^2} + \frac{1}{x}} \right) = - \infty ;$
$\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0 + 0} \left( {{x^2} + \frac{1}{x}} \right) = + \infty .$

Hence, $$x = 0$$ (the $$y-$$axis) is a vertical asymptote.

Now let's evaluate the following two limits:

$\lim\limits_{x \to \pm \infty } f\left( x \right) = \lim\limits_{x \to \pm \infty } \left( {{x^2} + \frac{1}{x}} \right) = + \infty ;$
$k = \lim\limits_{x \to \pm \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{{x^2} + \frac{1}{x}}}{x} = \lim\limits_{x \to \pm \infty } \left( {x + \frac{1}{{{x^2}}}} \right) = \pm \infty .$

As both these limits approach the infinity, we conclude that the function has neither horizontal, nor oblique asymptotes.

Take the derivative:

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

Find the critical points:

$f^\prime\left( x \right) = 0,\;\; \Rightarrow 2x - \frac{1}{{{x^2}}} = 0,\;\; \Rightarrow \frac{{2{x^3} - 1}}{{{x^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {{x^3} = \frac{1}{2}}\\ {{x^2} \ne 0} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = \frac{1}{{\sqrt{2}}}}\\ {x \ne 0} \end{array}} \right.,\;\; \Rightarrow x = \frac{1}{{\sqrt{2}}} \approx 0.79$

As you can see from the sign diagram, this point is a point of local minimum. Its $$y-$$coordinate is equal

$f\left( {\frac{1}{{\sqrt{2}}}} \right) = {\left( {\frac{1}{{\sqrt{2}}}} \right)^2} + \frac{1}{{\frac{1}{{\sqrt{2}}}}} = \frac{1}{{\sqrt{4}}} + \sqrt{2} = \frac{{1 + \sqrt{2} \cdot \sqrt{4}}}{{\sqrt{4}}} = \frac{{1 + \sqrt{8}}}{{\sqrt{4}}} = \frac{{1 + 2}}{{\sqrt{4}}} = \frac{3}{{\sqrt{4}}} \approx 1.89$

Thus, the function has a local minimum at $$\left( {\frac{1}{{\sqrt{2}}},\frac{3}{{\sqrt{4}}}} \right) = \left( {0.79,1.89} \right).$$

Now we compute the second derivative:

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

Determine the points where the $$2$$nd derivative is zero:

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

Using the sign chart (see above) we find that $$x = -1$$ is a point of inflection. This point coincides with the root of the function.

A schematic view of the function is shown in Figure $$7b$$ above.

### Example 8.

$y = {x^3}{e^x}.$

Solution.

The function is defined and differentiable on the whole real line. In this case, it does not have a vertical asymptote. Check for the existence of oblique (slant) asymptotes. Compute the following limits:

$\lim\limits_{x \to + \infty } \left( {{x^3}{e^x}} \right) = + \infty ;$
$\lim\limits_{x \to - \infty } \left( {{x^3}{e^x}} \right) = \lim\limits_{x \to - \infty } \frac{{\left( { - {{\left( { - x} \right)}^3}} \right)}}{{{e^{ - x}}}} = \left[ {\begin{array}{*{20}{l}} { - x = z,}\\ {x \to - \infty, }\\ {z \to + \infty } \end{array}} \right] = - \lim\limits_{z \to + \infty } \frac{{{z^3}}}{{{e^z}}} = - \lim\limits_{z \to + \infty } \frac{{3{z^2}}}{{{e^z}}} = - \lim\limits_{z \to + \infty } \frac{{6z}}{{{e^z}}} = - \lim\limits_{z \to + \infty } \frac{6}{{{e^z}}} = 0.$

When calculating the second limit we made the change $$\left( { - x} \right) \to z$$ and used L'Hopital's Rule. It can be seen that the function approaches zero as $$x \to -\infty,$$ that is the graph has the horizontal asymptote $$y = 0.$$

Calculate the roots of the function:

$y\left( x \right) = 0,\;\; \Rightarrow {x^3}{e^x} = 0,\;\; \Rightarrow {x^3} = 0,\;\; \Rightarrow x = 0.$

It follows that the function is positive for $$x \gt 0$$ and negative for $$x \lt 0$$ (Figure $$4a$$).

Find the first derivative and determine the extreme points and intervals of monotonicity:

$y'\left( x \right) = {\left( {{x^3}{e^x}} \right)^\prime } = {\left( {{x^3}} \right)^\prime }{e^x} + {x^3}{\left( {{e^x}} \right)^\prime } = 3{x^2}{e^x} + {x^3}{e^x} = {x^2}{e^x}\left( {3 + x} \right).$

Then the roots of the derivative have the following values:

$y'\left( x \right) = 0,\;\; \Rightarrow {x^2}{e^x}\left( {3 + x} \right) = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = - 3.$

The intervals of constant sign of the first derivative are indicated in Figure $$8a.$$ The function is strictly decreasing in the interval $$\left( { - \infty , - 3} \right)$$ and strictly increasing in the intervals $$\left( { -3,0} \right)$$ and $$\left( {0, + \infty} \right).$$ Thus, the point $$x = -3$$ is a minimum. The other critical point $$x = 0$$ is not a local extremum as when passing through this point the derivative does not change its sign. At the point of minimum we have

$y\left( { - 3} \right) = {\left( { - 3} \right)^3}{e^{ - 3}} \approx - 27 \cdot 0,0498 \approx - 1,34.$

Now we investigate the second derivative:

$y^{\prime\prime}\left( x \right) = \left[ {{x^2}{e^x}\left( {3 + x} \right)} \right]^\prime = \left[ {{e^x}\left( {3{x^2} + {x^3}} \right)} \right]^\prime = {\left( {{e^x}} \right)^\prime }\left( {3{x^2} + {x^3}} \right) + {e^x}{\left( {3{x^2} + {x^3}} \right)^\prime } = {e^x}\left( {3{x^2} + {x^3}} \right) + {e^x}\left( {6x + 3{x^2}} \right) = {e^x}\left( {{x^3} + 6{x^2} + 6x} \right) = x{e^x}\left( {{x^2} + 6x + 6} \right).$

Compute its roots:

$y^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow x{e^x}\left( {{x^2} + 6x + 6} \right) = 0,\;\; \Rightarrow {x^2} + 6x + 6 = 0,\;\; \Rightarrow D = 36 - 4 \cdot 6 = 12,\;\; \Rightarrow {x_{1,2}} = \frac{{ - 6 \pm \sqrt {12} }}{2} = - 3 \pm \sqrt 3 \; \approx - 4,73;\; - 1,27.$

So the second derivative has the following roots:

${x_1} = - 3 - \sqrt 3 ,\;\;{x_2} = - 3 + \sqrt 3 ,\;\;{x_3} = 0.$

When passing through each of these points the sign of the derivative is reversed (Figure $$8a$$). Therefore, these points are points of inflection. The approximate values of the corresponding coordinates $$y$$ are given by

$y\left( { - 3 - \sqrt 3 } \right) = {\left( { - 3 - \sqrt 3 } \right)^3}{e^{ - 3 - \sqrt 3 }} \approx - 0,93;$
$y\left( { - 3 + \sqrt 3 } \right) = {\left( { - 3 + \sqrt 3 } \right)^3}{e^{ - 3 + \sqrt 3 }} \approx - 0,57;$
$y\left( 0 \right) = 0.$

Now we can sketch the graph of the function (Figure $$8b$$).

### Example 9.

$f\left( x \right) = \frac{{{e^x}}}{x}.$

Solution.

The function is defined for all $$x$$ except the point $$x = 0$$ where it has a discontinuity.

The function has no roots, it is positive for $$x \gt 0$$ and negative for $$x \lt 0.$$

The function is neither even, nor odd.

Make sure it has a vertical asymptote at $$x = 0:$$

$\lim\limits_{x \to 0 - 0} f\left( x \right) = \lim\limits_{x \to 0 - 0} \frac{{{e^x}}}{x} = - \infty ;$
$\lim\limits_{x \to 0 + 0} f\left( x \right) = \lim\limits_{x \to 0 + 0} \frac{{{e^x}}}{x} = + \infty .$

Hence, $$x = 0$$ ($$y-$$axis) is a vertical asymptote.

Check for horizontal asymptotes. Using L'Hopital's rule, we have:

$\lim\limits_{x \to \pm \infty } f\left( x \right) = \lim\limits_{x \to \pm \infty } \frac{{{e^x}}}{x} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \pm \infty } \frac{{\left( {{e^x}} \right)^\prime}}{{\left( x \right)^\prime}} = \lim \limits_{x \to \pm \infty } \frac{{{e^x}}}{1} = \left\{ {\begin{array}{*{20}{l}} \infty\text{ as }x \to \infty \\ 0\text{ as }x \to -\infty \end{array}} \right..$

We see that $$y = 0$$ ($$x-$$axis) is a horizontal asymptote of the function as $$x \to -\infty.$$

Calculate the derivative:

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

Determine the critical points:

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

It follows from the sign chart that $$x = 1$$ is a point of local minimum. Its $$y-$$coordinate is

$f(1) = \frac{{{e^1}}}{1} = e,$

so the local minimum is at $$\left( {1,e} \right).$$

Next we are to investigate the concavity of the function using the second derivative:

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

Note that the quadratic function $${{x^2} - 2x + 2}$$ in the numerator does not have roots and is always positive. As the exponential function $${{e^x}}$$ is also positive everywhere, then the sign of the second derivative depends only on the sign of the denominator. Hence, the function is concave downward on $$\left( { - \infty ,0} \right)$$ and concave upward on $$\left( {0, +\infty} \right).$$

A schematic graph of the function is given above.

### Example 10.

$y = {x^2}{e^{\frac{1}{x}}}.$

Solution.

The function is defined for all real values of $$x,$$ except the point $$x = 0,$$ where there is a discontinuity. Find one-sided limits at this point:

$\lim\limits_{x \to 0 - 0} y\left( x \right) = \lim\limits_{x \to 0 - 0} \left( {{x^2}{e^{\frac{1}{x}}}} \right) = 0;$
$\lim\limits_{x \to 0 + 0} y\left( x \right) = \lim\limits_{x \to 0 + 0} \left( {{x^2}{e^{\frac{1}{x}}}} \right) = \lim\limits_{x \to 0 + 0} \frac{{{e^{\frac{1}{x}}}}}{{{{\left( {\frac{1}{x}} \right)}^2}}} = \left[ {\begin{array}{*{20}{l}} {\frac{1}{x} = z,}\\ {x \to 0 + ,}\\ {z \to + \infty } \end{array}} \right] = \lim\limits_{z \to + \infty } \frac{{{e^z}}}{{{z^2}}} = \lim\limits_{z \to + \infty } \frac{{{e^z}}}{{2z}} = \lim\limits_{z \to + \infty } \frac{{{e^z}}}{2} = + \infty .$

When calculating the second limit we made replacement $${{\frac{1}{x}} \to z}$$ and used L'Hopital's Rule.

Thus, the line $$x = 0$$ (i.e. $$y$$-axis) is a vertical asymptote of this function. The function is always positive, except at the point $$x = 0,$$ in which the left-hand limit is $$y\left( { - 0} \right) = 0.$$ We also need to check for slant asymptotes as $$x \to \pm \infty:$$

$k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{{x^2}{e^{\frac{1}{x}}}}}{x} = \lim\limits_{x \to \pm \infty } \left( {x{e^{\frac{1}{x}}}} \right) = \lim\limits_{x \to \pm \infty } \frac{{{e^{\frac{1}{x}}}}}{{\frac{1}{x}}} = \left[ {\begin{array}{*{20}{l}} {\frac{1}{x} = z,}\\ {x \to \pm \infty ,}\\ {z \to 0} \end{array}} \right] = \lim\limits_{z \to 0} \frac{{{e^z}}}{z} = \infty .$

Consequently, the function has no oblique asymptotes.

Calculate the first derivative and determine the stationary points:

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

The derivative is negative to the left of $$x = {\frac{1}{2}}$$, and positive to the right (Figure $$10a$$). So $$x = {\frac{1}{2}}$$ is a local minimum. The minimum value is

$y\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^2}{e^{\frac{1}{{\frac{1}{2}}}}} = \frac{1}{4}{e^2} \approx 1,85.$

The second derivative is

$y^{\prime\prime}\left( x \right) = \left[ {{e^{\frac{1}{x}}}\left( {2x - 1} \right)} \right]^\prime = {\left( {{e^{\frac{1}{x}}}} \right)^\prime }\left( {2x - 1} \right) + {e^{\frac{1}{x}}}{\left( {2x - 1} \right)^\prime } = {e^{\frac{1}{x}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right)\left( {2x - 1} \right) + 2{e^{\frac{1}{x}}} = {e^{\frac{1}{x}}}\left( {2 - \frac{{2x - 1}}{{{x^2}}}} \right) = {e^{\frac{1}{x}}}\frac{{2{x^2} - 2x + 1}}{{{x^2}}}.$

In the last expression, the quadratic function in the numerator does not have real roots and is always positive. Given that the denominator contains $${x^2},$$ we conclude that the second derivative is positive for all $$x \ne 0.$$ Therefore, the function is strictly convex downward in the intervals $$\left( { - \infty ,0} \right)$$ and $$\left( {0, +\infty} \right).$$ This means that the function has no inflection points.

A schematic graph of the function is shown in Figure $$10b$$.

### Example 11.

$f\left( x \right) = x\ln x.$

Solution.

The function is defined for $$x \gt 0.$$ Determine the $$x-$$intercepts:

$f\left( x \right) = 0,\;\; \Rightarrow x\ln x = 0,\;\; \Rightarrow {x} = 1.$

The other root $$x = 0$$ does not belong to the domain of the function.

Clearly that the function is negative for $$0 \lt x \lt 1$$ and positive for $$x \gt 1.$$

Look for the vertical asymptote at $$x = 0.$$ Using L’Hopital’s rule, we can write the right-sided limit in the form

$\lim\limits_{x \to 0 + 0} \left( {x\ln x} \right) = \lim\limits_{x \to 0 + 0} \frac{{\ln x}}{{\frac{1}{x}}} = \lim\limits_{x \to 0 + 0} \frac{{\left( {\ln x} \right)^\prime}}{{\left( {\frac{1}{x}} \right)^\prime}} = \lim\limits_{x \to 0 + 0} \frac{{\frac{1}{x}}}{{ - \frac{1}{{{x^2}}}}} = - \lim\limits_{x \to 0 + 0} x = 0.$

The function has neither vertical, nor horizontal asymptotes. Consider the possible oblique asymptote as $$x \to \infty:$$

$k = \lim\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \infty } \frac{{\cancel{x}\ln x}}{\cancel{x}} = \lim\limits_{x \to \infty } \ln x = \infty .$

So there are no oblique asymptotes.

Take the derivative:

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

Find the critical points:

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

The derivative changes sign from negative to positive at $$x = \frac{1}{e}.$$ Therefore, $$x = \frac{1}{e}$$ is a point of local minimum. The minimum value is

${f_{\min }} = f\left( {\frac{1}{e}} \right) = \frac{1}{e}\ln \frac{1}{e} = \frac{1}{e} \cdot \left( { - 1} \right) = - \frac{1}{e}.$

Hence the minimum point is at $$\left( {\frac{1}{e}, - \frac{1}{e}} \right).$$

Calculate the second derivative:

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

Since $$f^{\prime\prime}\left( x \right) \gt 0$$ for all $$x$$ in the domain, the function is concave up everywhere (Figure $$11b$$).

### Example 12.

$y = \frac{{{x^2} - 1}}{{{x^3}}}.$

Solution.

The function is defined for all real $$x,$$ except the point $$x = 0.$$ Calculate one-sided limits at this point:

$\lim\limits_{x \to 0 - 0} \frac{{{x^2} - 1}}{{{x^3}}} = + \infty ,\;\;\lim\limits_{x \to 0 + 0} \frac{{{x^2} - 1}}{{{x^3}}} = - \infty .$

Hence, the line $$x = 0$$ ($$y$$-axis) is a vertical asymptote. We also check for the existence of slant and horizontal asymptotes:

$k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{\frac{{{x^2} - 1}}{{{x^3}}}}}{x} = \lim\limits_{x \to \pm \infty } \frac{{{x^2} - 1}}{{{x^4}}} = \lim\limits_{x \to \pm \infty } \frac{{\frac{1}{{{x^2}}} - \frac{1}{{{x^4}}}}}{1} = 0;$
$b = \lim\limits_{x \to \pm \infty } \left[ {y\left( x \right) - kx} \right] = \lim\limits_{x \to \pm \infty } \left[ {\frac{{{x^2} - 1}}{{{x^3}}} - 0} \right] = \lim\limits_{x \to \pm \infty } \frac{{\frac{1}{x} - \frac{1}{{{x^3}}}}}{1} = 0.$

Thus, the graph of this function also has a horizontal asymptote given by the equation $$y = 0.$$

Note that the function is odd since

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

Determine the intersection points of the graph with the $$x$$-axis:

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

The function takes positive values on the intervals $$\left( { - 1,0} \right)$$ and $$\left( {1, +\infty} \right)$$ and negative values on the intervals $$\left( {-\infty, -1} \right)$$ and $$\left( {0,1} \right)$$ (Figure $$12a$$).

Calculate the derivative:

$y'\left( x \right) = \left( {\frac{{{x^2} - 1}}{{{x^3}}}} \right)^\prime = {\left( {\frac{1}{x} - \frac{1}{{{x^3}}}} \right)^\prime } = {\left( {{x^{ - 1}} - {x^{ - 3}}} \right)^\prime } = - {x^{ - 2}} + 3{x^{ - 4}} = - \frac{1}{{{x^2}}} + \frac{3}{{{x^4}}} = \frac{{3 - {x^2}}}{{{x^4}}}.$

Find the stationary points:

$y'\left( x \right) = 0,\;\; \Rightarrow \frac{{3 - {x^2}}}{{{x^4}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {3 - {x^2} = 0}\\ {{x^4} \ne 0} \end{array}} \right.,\;\; \Rightarrow {x_{1,2}} = \pm \sqrt 3 \approx 1,73.$

The point $$x = - \sqrt 3$$ is a minimum point as when passing through it the derivative changes sign from minus to plus (Figure $$12a$$). The other point $$x = \sqrt 3$$ is a maximum. The function has the following value at the first point:

$y\left( { - \sqrt 3 } \right) = \frac{{{{\left( { - \sqrt 3 } \right)}^2} - 1}}{{{{\left( { - \sqrt 3 } \right)}^3}}} = - \frac{2}{{3\sqrt 3 }} \approx - 0,38.$

By virtue of the oddness of the function, we can immediately write:

$y\left( {\sqrt 3 } \right) = \frac{2}{{3\sqrt 3 }} \approx 0,38.$

Find the second derivative of the function:

$y^{\prime\prime}\left( x \right) = {\left( { - {x^{ - 2}} + 3{x^{ - 4}}} \right)^\prime } = 2{x^{ - 3}} - 12{x^{ - 5}} = \frac{2}{{{x^3}}} - \frac{{12}}{{{x^5}}} = \frac{{2{x^2} - 12}}{{{x^5}}} = \frac{{2\left( {{x^2} - 6} \right)}}{{{x^5}}}.$

The second derivative is zero at the following points:

$y^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{2\left( {{x^2} - 6} \right)}}{{{x^5}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{x^2} - 6 = 0}\\ {{x^5} \ne 0} \end{array}} \right.,\;\; \Rightarrow {x_{1,2}} = \pm \sqrt 6 \approx \pm 2,45.$

It follows from Figure $$12a$$ that the function is strictly convex upward in the intervals $$\left( { - \infty , - \sqrt 6 } \right)$$ and $$\left( {0, \sqrt 6 } \right)$$ and strictly convex downward in the intervals $$\left( {-\sqrt 3, 0 } \right)$$ and $$\left( {\sqrt 6, +\infty } \right).$$ When passing through the points $$x = - \sqrt 3$$ and $$x = \sqrt 3$$ the second derivative changes its sign. Hence, these are the points of inflection. Calculate the values of the function at these points:

$y\left( { - \sqrt 6 } \right) = \frac{{{{\left( { - \sqrt 6 } \right)}^2} - 1}}{{{{\left( { - \sqrt 6 } \right)}^3}}} = - \frac{5}{{6\sqrt 6 }} \approx - 0,34.$

Again, since the function is odd, we obtain:

$y\left( {\sqrt 6 } \right) = \frac{5}{{6\sqrt 6 }} \approx 0,34.$

Now we have enough information to sketch a graph of the function (Figure $$12b$$).

### Example 13.

$y = \sqrt{{{x^2}\left( {x + 1} \right)}}.$

Solution.

The function is defined on the entire real axis. Its graph crosses the $$x$$-axis at $$x = 0$$ and $$x = -1.$$ The function is positive on the intervals $$\left( { - 1,0} \right)$$ and $$\left( {0, +\infty} \right)$$ and negative on the interval $$\left( {-\infty, -1} \right)$$ (Figure $$13a$$).

Since the function is continuous everywhere, it does not have a vertical asymptote. We calculate the coefficients of the oblique (slant) asymptote:

$k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} = \lim\limits_{x \to \pm \infty } \frac{{\sqrt{{{x^2}\left( {x + 1} \right)}}}}{x} = \lim\limits_{x \to \pm \infty } \sqrt{{\frac{{{x^3} + {x^2}}}{{{x^3}}}}} = \lim\limits_{x \to \pm \infty } \sqrt{{\frac{{1 + \frac{1}{x}}}{1}}} = 1;$
$b = \lim\limits_{x \to \pm \infty } \left[ {y\left( x \right) - kx} \right] = \lim\limits_{x \to \pm \infty } \left( {\sqrt{{{x^2}\left( {x + 1} \right)}} - x} \right) = \lim\limits_{x \to \pm \infty } \left( {\sqrt{{{x^3} + {x^2}}} - \sqrt{{{x^3}}}} \right) = \frac{1}{3}.$

Consequently, there is an oblique asymptote given by the equation $$y = x + {\frac{1}{3}}.$$

Find the derivative:

$y'\left( x \right) = {\left[ {\sqrt{{{x^2}\left( {x + 1} \right)}}} \right]^\prime } = {\left[ {{{\left( {{x^3} + {x^2}} \right)}^{\frac{1}{3}}}} \right]^\prime } = \frac{1}{3}{\left( {{x^3} + {x^2}} \right)^{ - \frac{2}{3}}} \cdot \left( {3{x^2} + 2x} \right) = \frac{{3{x^2} + 2x}}{{3\sqrt{{{x^4}{{\left( {x + 1} \right)}^2}}}}} = \frac{{3x + 2}}{{3\sqrt{{x{{\left( {x + 1} \right)}^2}}}}}.$

It follows from here that the function has the following critical points:

${x_1} = 0,\;\;{x_2} = - 1,\;\;{x_3} = - \frac{2}{3}.$

The derivative does not exist at the first two points. However, when passing through the point $$x =0,$$ the derivative changes sign from minus to plus (Figure $$13a$$). Hence, there is a minimum at this point (the cusp point). The value of the function value at $$x =0$$ has already been found above: $$y\left( 0 \right) = 0.$$ There is no extreme point at $$x = -1,$$ because when passing through it the derivative does not change its sign. At the point $$x = - {\frac{2}{3}}$$ the derivative is zero and changes sign from plus to minus when passing through this point. It is clear that the function has a maximum at $$x = - {\frac{2}{3}}$$ equal

$y\left( { - \frac{2}{3}} \right) = \sqrt{{{{\left( { - \frac{2}{3}} \right)}^2} \cdot \left( { - \frac{2}{3} + 1} \right)}} = \sqrt{{\frac{4}{9} \cdot \frac{1}{3}}} = \sqrt{{\frac{4}{{27}}}} = \frac{{\sqrt{4}}}{3} \approx 0,53.$

Now we investigate the possible inflection points and convexity of the function. The second derivative is given by

$y^{\prime\prime}\left( x \right) = {\left[ {\frac{1}{3}{{\left( {{x^3} + {x^2}} \right)}^{ - \frac{2}{3}}}\left( {3{x^2} + 2x} \right)} \right]^\prime } = {\frac{1}{3} \left[ {\frac{{6x + 2}}{{{{\left( {{x^3} + {x^2}} \right)}^{\frac{2}{3}}}}} - \frac{{2{x^2}{{\left( {3x + 2} \right)}^2}}}{{3{{\left( {{x^3} + {x^2}} \right)}^{\frac{5}{3}}}}}} \right] } = - \frac{{2{x^2}}}{{9{x^{\frac{{10}}{3}}}{{\left( {x + 1} \right)}^{\frac{5}{3}}}}} = - \frac{2}{{9\sqrt{{{x^4}{{\left( {x + 1} \right)}^5}}}}}.$

Thus, the second derivative is nowhere zero. However, the singular point $$x = -1$$ is an inflection point, as when passing through it, the sign of the second derivative is reversed (Figure $$13a$$). In contrast, the other singular point $$x = 0$$ is not an inflection point. The function is convex downward on the interval $$\left( { - \infty , - 1} \right)$$ and convex upward on the intervals $$\left( { - 1,0} \right)$$ and $$\left( {0, +\infty} \right).$$

Considering all the above points, we can draw the graph of the function (Figure $$13b$$).

### Example 14.

$y = {x^2}\sqrt {x + 1} .$

Solution.

It is clear that the function is defined for $$x \ge -1$$ and is non-negative in its domain. The function has no asymptotes.

We find the points of intersection of the graph with the coordinate axes:

$y\left( 0 \right) = {0^2}\sqrt {0 + 1} = 0;$
$y\left( x \right) = 0,\;\; \Rightarrow {x^2}\sqrt {x + 1} = 0,\;\; \Rightarrow {x_1} = 0,\;{x_2} = - 1.$

Consequently, there are the following points of intersection: $$\left( {0,0} \right)$$ and $$\left( {-1,0} \right).$$

Calculate the derivative:

$y'\left( x \right) = {\left( {{x^2}\sqrt {x + 1} } \right)^\prime } = 2x \cdot \sqrt {x + 1} + {x^2} \cdot \frac{1}{{2\sqrt {x + 1} }} = 2x\sqrt {x + 1} + \frac{{{x^2}}}{{2\sqrt {x + 1} }} = \frac{{4x\left( {x + 1} \right) + {x^2}}}{{2\sqrt {x + 1} }} = \frac{{5{x^2} + 4x}}{{2\sqrt {x + 1} }}.$

The critical points have the following values:

${x_1} = - 1,\;\;{x_2} = 0,\;\;{x_3} = - \frac{4}{5}.$

The function is increasing on the intervals $$\left( { - 1, - {\frac{4}{5}}} \right)$$ and $$\left( {0, + \infty } \right)$$ and is decreasing on the interval $$\left( {- {\frac{4}{5}}, 0} \right).$$ At the point $$x = - {\frac{4}{5}},$$ the function has a maximum equal

$y\left( { - \frac{4}{5}} \right) = {\left( { - \frac{4}{5}} \right)^2}\sqrt { - \frac{4}{5} + 1} = \frac{{16}}{{25\sqrt 5 }} \approx 0,28.$

There is also a minimum at the point $$x=0$$ so that $$y\left( 0 \right) = 0.$$

Consider the second derivative:

$y^{\prime\prime}\left( x \right) = {\left( {\frac{{5{x^2} + 4x}}{{2\sqrt {x + 1} }}} \right)^\prime } = \frac{{\left( {20x + 8} \right)\left( {x + 1} \right) - \left( {5{x^2} + 4x} \right)}}{{4\sqrt {{{\left( {x + 1} \right)}^3}} }} = \frac{{\color{blue}{20{x^2}} + \color{red}{8x} + \color{red}{20x} + \color{green}{8} - \color{blue}{5{x^2}} - \color{red}{4x}}}{{4\sqrt {{{\left( {x + 1} \right)}^3}} }} = \frac{{\color{blue}{15{x^2}} + \color{red}{24x} + \color{green}{8}}}{{4\sqrt {{{\left( {x + 1} \right)}^3}} }}.$

The second derivative is zero at the following points:

$y^{\prime\prime}\left( x \right) = 0,\;\; \Rightarrow \frac{{15{x^2} + 24x + 8}}{{4\sqrt {{{\left( {x + 1} \right)}^3}} }} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {15{x^2} + 24x + 8 = 0}\\ {x \ne - 1} \end{array}} \right.,\;\; \Rightarrow D = 576 - 4 \cdot 15 \cdot 8 = 96,\;\; \Rightarrow {x_{1,2}} = \frac{{ - 24 \pm \sqrt {96} }}{{30}} = \frac{{ - 24 \pm 4\sqrt 6 }}{{30}} = \frac{2}{{15}}\left( { - 6 \pm \sqrt 6 } \right) \approx - 1,13;\; - 0,47.$

One of the roots

$x = {\frac{2}{{15}}}\left( { - 6 - \sqrt 6 } \right) \approx - 1,13$

lies outside of the domain. When passing through the second root

$x = {\frac{2}{{15}}}\left( { - 6 + \sqrt 6 } \right) \approx - 0,47$

the second derivative changes sign (Figure $$14a$$). Therefore, this point is a point of inflection. The value of the function here is approximately equal to

$y\left( {\frac{2}{{15}}\left( { - 6 + \sqrt 6 } \right)} \right) \approx y\left( { - 0,47} \right) \approx 0,16.$

The graph of the function is convex upward to the left of the indicated point, and is convex downward to the right. A schematic view of the function is shown in Figure $$14b$$.