Calculus

Applications of the Derivative

Applications of Derivative Logo

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:

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

Solve the quadratic equation:

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

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 \(5a\)).

Sign chart for the function f(x)=(x^2+1)/(x-1).
Figure 5a.
Graph of the function f(x)=(x^2+1)/(x-1).
Figure 5b.

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 \(6b\).

Graph of the function f(x)=1/(1-x^2).
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:

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

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

\[f\left( {\frac{1}{{\sqrt[3]{2}}}} \right) = {\left( {\frac{1}{{\sqrt[3]{2}}}} \right)^2} + \frac{1}{{\frac{1}{{\sqrt[3]{2}}}}} = \frac{1}{{\sqrt[3]{4}}} + \sqrt[3]{2} = \frac{{1 + \sqrt[3]{2} \cdot \sqrt[3]{4}}}{{\sqrt[3]{4}}} = \frac{{1 + \sqrt[3]{8}}}{{\sqrt[3]{4}}} = \frac{{1 + 2}}{{\sqrt[3]{4}}} = \frac{3}{{\sqrt[3]{4}}} \approx 1.89\]
Sign chart for the function f(x)=x^2+1/x.
Figure 7a.
Graph pf the function f(x)=x^2+1/x.
Figure 7b.

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

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 \(8b\)).

Signs of the derivatives of the function y=x^3*exp(x).
Figure 8a.
Graph of the function y=x^3*exp(x).
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:

\[f^\prime\left( x \right) = 0,\;\; \Rightarrow \frac{{{e^x}\left( {x - 1} \right)}}{{{x^2}}} = 0,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1}\\ {x \ne 0} \end{array}} \right.,\;\; \Rightarrow x = 1.\]
Sign chart of the function f(x)=exp(x)/x.
Figure 9a.
Graph of the function f(x)=exp(x)/x.
Figure 9b.

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 \(10b\).

Signs of the derivatives of the function y=x^2*exp(1/x).
Figure 10a.
Graph of the function y=x^2*exp(1/x).
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 \(11b\)).

Graph of the function f(x)=x*ln(x).
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 \(12b\)).

Signs of the derivatives of the rational function y=(x^2-1)/x^3.
Figure 12a.
A schematic view of the rational function y=(x^2-1)/x^3.
Figure 12b.

Example 13.

\[y = \sqrt[3]{{{x^2}\left( {x + 1} \right)}}.\]

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:

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

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

Find the derivative:

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

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

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

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

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

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 \(13b\)).

Signs of the derivatives of a cubic root function.
Figure 13a.
A schematic view of a cubic root 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 \(14b\).

Signs of the derivative of the function y=x^2*sqrt(x+1).
Figure 14a.
Graph of the function y=x^2*sqrt(x+1).
Figure 14b.
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