7.7 Graphs II: Elementary Functions

7.7.2 Inverse Trigonomic Functions

Example 1
Given the function \(f\) together with its first and second derivatives,
\[f(x) =\arcsin \left ( \frac{1-x^2}{1+ x^2}\right ),\;\;\; f'(x) = – \frac{2x}{|x|(x^2+1)}, \;\;\; f”(x)=\frac{4x^2}{|x|(x^2+1)^2}, \]
sketch its graph by completing steps (I)–(VI).

(I) Domain and Intercepts
\(D_f=(-\infty, \infty)\). \(y\)-intercept: \(\displaystyle \left (0,f(0)\right ) = \left (0,\frac{\pi}{2}\right )\).

\(x\)-intercepts: \(\displaystyle f(x) = 0 \Longleftrightarrow \frac{1-x^2}{1+ x^2} = 0 \Longleftrightarrow x=\pm 1. \Longrightarrow \{(\pm 1, 0)\}\).

(II) Symmetry

The function is even, since \(f(-x)=f(x)\) (explain why). Thus, the graph is symmetric with respect to the \(y\)-axis.

(III) Asymptotic Behavior

The graph has no vertical asymptotes.

Horizontal asymptotes. By continuity of \(y=\arcsin x\), we have in \((*)\)
\(\lim_{x\to \pm \infty} \arcsin \left ( \frac{1-x^2}{1+ x^2}\right ) \underset{(*)}{=} \arcsin\lim_{x\to \pm \infty}\frac{1-x^2}{1+x^2} = \arcsin (-1) = -\frac{\pi}{2}. \)
Hence, the graph has a horizontal asymptote given by \(\displaystyle y=-\frac{\pi}{2}\).

(IV) Monotonicity and Local Extrema

Since
\( f'(x) = – \frac{2x}{|x|(x^2+1)} = \left [\renewcommand{\arraystretch}{2}\begin{array}{ll}
\displaystyle – \frac{2 }{ x^2+1} & \;\text{if } \; x > 0\\
\displaystyle \frac{2 }{ x^2+1} & \;\text{if } \; x < 0
\end{array}\right .,
\)
\(f’\) is undefined at \(x=0\) and vanishes nowhere. This,\(\displaystyle \left \{0\right \}\) is its only critical number.
It is clear that \(f'(x) < 0 \) if \(x > 0\) and \(f'(x) > 0 \) if \(x < 0 \).

Thus, \(f\) is increasing on \((-\infty, 0)\), and decreasing on \((0, \infty)\).

It follows that \(f\) has a local maximum value \(\displaystyle f(0)=\frac{\pi}{2}\) and no local minimum values.

(V) Concavity and Inflection Points

Since
\( f”(x) =\frac{4x^2}{|x|(x^2+1)^2} = \left [\renewcommand{\arraystretch}{2}\begin{array}{ll}
\displaystyle \frac{4x }{ x^2+1} & \;\text{if } \; x > 0\\
\displaystyle -\frac{4x }{ x^2+1} & \;\text{if } \; x < 0
\end{array}\right .,
\)
clearly, \(f”(x) > 0 \) if \(x < 0\) and \(f”(x) < 0 \) if \(x > 0 \).

Thus, \(f\) is concave upward on \((-\infty, 0)\), and concave downward on \((0, \infty)\), and has no inflection points.

(VI) Graph

Based on all the above information, the graph is as shown in Figure 7.38.

Figure 7.38 Example 1

Example 2
Given the function \(f\) together with its first and second derivatives,
\(f(x) =\arctan \left (e^{-2x}-1\right ) ,\;\;\; f'(x) = – \frac{2e^{-2x}}{ \left (e^{-2x}-1\right )^2+1}, \;\;\; f”(x)=-\frac{4 e^{-2 x} \left (e^{-4 x}-2\right )}{\left ( \left (e^{-2x}-1\right )^2+1\right )^2}, \)
sketch its graph by completing steps (I)–(VI).

(I) Domain and Intercepts

\(D_f=(-\infty, \infty)\). \(y\)-intercept: \(\displaystyle \left (0,f(0)\right ) = \left (0,0\right )\).

\(x\)-intercepts: \(\displaystyle f(x) = 0 \Longleftrightarrow e^{-2x}-1= 0 \Longleftrightarrow x=0. \Longrightarrow \{(0, 0)\}\).

(II) Symmetry

No symmetry since \(f\) is neither even nor odd (explain).

(III) Asymptotic Behavior

The graph has no vertical asymptotes.

Horizontal asymptotes. By continuity of \(y=\arctan x\), we have in \((*)\)
\(\lim_{x\to \infty} \arctan\left (e^{-2x}-1\right )\underset{(*)}{=} \arctan\left (\lim_{x\to \infty}\left (e^{-2x}-1\right )\right ) = \arctan (-1) = -\frac{\pi}{4}. \)
\(\lim_{x\to – \infty} \arctan\left (e^{-2x}-1\right )\underset{(*)}{=} \arctan\left (\lim_{x\to – \infty}\left (e^{-2x}-1\right )\right ) = -\frac{\pi}{2}. \)

Hence, the graph has a horizontal asymptotes \(\displaystyle y=-\frac{\pi}{4}\) and \(\displaystyle y=\frac{\pi}{2}\).

(IV) Monotonicity and Local Extrema

Since
\(\displaystyle f'(x) = – \frac{2e^{-2x}}{ \left (e^{-2x}-1\right )^2+1}\), \(f'(x) < 0 \) for all \(x\in \mathbb{R}\). Thus, \(f\) is decreasing on \(\mathbb{R}\). Hence, it has no local maximum values and no local minimum values.

(V) Concavity and Inflection Points

\( f”(x) =0 \Longleftrightarrow -\frac{4 e^{-2 x} \left (e^{-4 x}-2\right )}{\left ( \left (e^{-2x}-1\right )^2+1\right )^2} = 0 \Longleftrightarrow e^{-4 x}-2=0 \Longleftrightarrow x= -\frac{1}{4}\ln 2. \)
Given that the denominator of \(f”\) and the factor \(4 e^{-2 x}\) in the numerator
are strictly positive, it follows that
\(f”(x) > 0 \Longleftrightarrow -\left (e^{-4 x}-2\right ) > 0 \Longleftrightarrow e^{-4 x}< 2 \Longleftrightarrow x > -\frac{1}{4}\ln 2 \)

Thus, \(f\) is concave upward on \(\displaystyle \left ( -\frac{1}{4}\ln 2 , \infty\right )\), and concave downward on \(\displaystyle \left (-\infty, -\frac{1}{4}\ln 2 \right )\).

\(f\) has an inflection point at \(\displaystyle \left ( -\frac{1}{4}\ln 2, f\left ( -\frac{1}{4}\ln 2\right )\right ) = \left ( -\frac{1}{4}\ln 2, \arctan\left (\sqrt{2}-1\right )\right )\) (explain).

(VI) Graph

Based on all the above information, the graph is as shown in Figure 7.39.

Figure 7.39 Example 2

7.7.3 Exponential Functions

Example 1
Given the function \(f\) together with its first and second derivatives,
\[f(x) =\frac{e^x}{1+e^{2x}},\;\;\; f'(x) = \frac{e^x – e^{3x}}{(1+e^{2x})^2}, \;\;\; f”(x) = \frac{e^x -6 e^{3x} + e^{5x}}{(1+e^{2x})^3}, \]
sketch its graph by completing steps (I)–(VI).

(I) Domain and Intercepts

\(D_f=(-\infty,\infty)\). \(x\)-intercepts: none; \(y\)-intercept: \(\displaystyle \left (0, \frac{1}{2}\right )\).

(II) Symmetry

The function is even
\[f(-x) = \frac{e^{-x}}{1+e^{-2x}} = \frac{e^{-x}}{1+e^{-2x}}\cdot \frac{e^{2x}}{e^{2x}} = \frac{e^{x}}{1+e^{2x}} = f(x).\]
Thus, the graph is symmetric with respect to the \(y\)-axis.

(III) Asymptotic Behavior

No vertical asymptotes.

Horizontal asymptotes:

\[\lim_{x\to \infty} \frac{e^{x}}{1+e^{2x}}\;\; \underset{\text{l’H}}{=}\;\;\lim_{x\to \infty} \frac{e^{x}}{2e^{2x}}=\lim_{x\to \infty} \frac{1}{2e^{x}}= 0, \]
and, by symmetry, \(\displaystyle \lim_{x\to -\infty} \frac{e^{x}}{1+e^{2x}} =0\).

Thus, the graph has a horizontal asymptotes: \(y = 0\).

(IV) Monotonicity and Local Extrema

Since \(\displaystyle f'(x) = \frac{e^x – e^{3x}}{(1+e^{2x})^2} = \frac{e^x(1 – e^{x})(1+e^x)}{(1+e^{2x})^2}\), \(f\) has only one critical number: \(\displaystyle \left \{0\right \}\).

Verify that the following is true: \(f\) is increasing on \((-\infty,0) \), and decreasing on \((0, \infty)\).

It follows that \(f\) has a local maxumum value \(\displaystyle f(0)=\frac{1}{2}\) and no local minimum values.

(V) Concavity and Inflection Points

Since
\[f”(x) = \frac{e^x -6 e^{3x} + e^{5x}}{(1+e^{2x})^3}= \frac{e^x \left ( e^{4x} -6e^{2x} +1\right )}{(1+e^{2x})^3},\]

\[
f”(x) = 0 \Longleftrightarrow
e^x \left ( e^{4x} -6e^{2x} +1\right ) =0 \Longleftrightarrow
\left [\renewcommand{\arraystretch}{1.2}\begin{array}{l}
u = e^{2x}\\
u^2-6u+1=0
\end{array}\right .
\Longleftrightarrow
\left [\renewcommand{\arraystretch}{1.2}\begin{array}{l}
u = e^{2x}\\
u = 3\pm 2\sqrt{2}
\end{array}\right . \Longleftrightarrow
\]

\[\Longleftrightarrow
e^{2x} = 3\pm 2\sqrt{2} \Longleftrightarrow 2x= \ln\left (3\pm 2\sqrt{2} \right )
\Longleftrightarrow x= \frac{1}{2}\ln\left (3\pm 2\sqrt{2} \right ).
\]

Given that

\[f”(x) = \frac{e^x \left ( e^{4x} -6e^{2x} +1\right )}{(1+e^{2x})^3} = \frac{e^x }{(1+e^{2x})^3}\cdot\left ( e^{4x} -6e^{2x} +1\right ), \;\; \text{ with } \;\; \frac{e^x }{(1+e^{2x})^3} > 0, \]

to determine the sign of \(f”\) it is only necessary to determine the sign of the factor

\[ e^{4x} -6e^{2x} +1 = \left ( e^{2x} -\left (3- 2\sqrt{2}\right )\right ) \left ( e^{2x} – \left (3+ 2\sqrt{2}\right )\right ).\]

Since \(y=e^{2x}\) is increasing, it follows that

(i) \(f”(x) > 0\), if \(\displaystyle x > \frac{1}{2}\ln\left (3 + 2\sqrt{2}\right )\) (both factors are positive);

(ii) \(f”(x) < 0\), if \(\displaystyle \frac{1}{2}\ln\left (3 + 2\sqrt{2}\right ) > x > \frac{1}{2}\ln\left (3 – 2\sqrt{2}\right ) \) (one factor positive the other negative);

(iii) \(f”(x) > 0\), if \(\displaystyle x < \frac{1}{2}\ln\left (3 – 2\sqrt{2}\right )\) (both factors are negative).

\(f\) is concave upward on \(\displaystyle \left (-\infty, \frac{1}{2}\ln\left (3 – 2\sqrt{2}\right )\right )\cup \left( \frac{1}{2}\ln\left (3 + 2\sqrt{2}\right ), \infty\right)\), and downward on \(\displaystyle \left ( \frac{1}{2}\ln\left (3 – 2\sqrt{2}\right ), \frac{1}{2}\ln\left (3 + 2\sqrt{2}\right )\right )\).

The inflection points are located at \(\displaystyle x = \frac{1}{2}\ln\left (3 \pm 2\sqrt{2}\right )\).

(VI) Graph

Based on all the above information, the graph is as shown in Figure~\ref{ch07-07-f08}.

Figure 7.40 Example 1

Trigonometry Review

The following example makes use of a basic procedure to rewrite the sum
\(
\hphantom{\hspace{.5in}(*)}
a\sin x +b\cos x = \sqrt{a^2+b^2}\,\sin (x + \theta),\;\; \;\; a^2+b^2 \neq 0,\; a,b\in \mathbb{R},\hspace{.5in}(*)
\)
where \(\theta\) satisfies
\(\cos \theta = \frac{a}{\displaystyle \sqrt{a^2+b^2}}, \; \; \sin \theta = \frac{b}{\displaystyle \sqrt{a^2+b^2}}. \)
To see that this is valid, note that

(i) the point \(\displaystyle \left (\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right )\) belongs to the unit circle (explain why), hence such a \(\theta\) exists;

(ii) by the addition formula for \(\sin (x + \theta)\), we have
\begin{align*}
a\sin x +b\cos x &= \sqrt{a^2+b^2}\left (\frac{a}{ \sqrt{a^2+b^2}}\sin x + \frac{b}{ \sqrt{a^2+b^2}}\cos x\right )=
\vphantom{\frac{A}{\frac{A}{A}}}\\
& = \sqrt{a^2+b^2}\left (\cos \theta \sin x + \sin \theta\sin x \right ) = \sqrt{a^2+b^2}\,\sin (x+\theta).
\end{align*}

Note that \(\theta\) is not unique,
\(a\sin x +b\cos x = \sqrt{a^2+b^2}\,\sin (x+\theta + 2\pi k), \; \forall k \in \mathbb{Z}. \)

Thus, there is the option to look for a value of \(\theta\) for which it may be simpler to work with. The following example shows how to use this to find critical numbers and inflection points.

Example 2. Given the function \(f\) together with its first and second derivatives,
\[f(x) =e^{x/3}\sin 2x ,\;\;\; f'(x) = \frac{1}{3}e^{x/3}\left (\sin 2x + 6 \cos 2x\right ), \;\;\; f”(x) = \frac{1}{9}e^{x/3}\left (-35\sin 2x + 12 \cos 2x\right ), \]
sketch its graph by completing steps (I)–(VI).

(I) Domain and Intercepts

\(D_f=(-\infty,\infty)\). \(y\)-intercept: \(\displaystyle \left (0, 0\right )\);

\(x\)-intercepts:
\begin{align*}
f(x) = 0 &\;\; \Longleftrightarrow \;\; e^{x/3}\sin 2x = 0 \;\; \Longleftrightarrow \;\; \sin 2x = 0 \;\; \Longleftrightarrow \;\; 2x = \pi k, \; k\in \mathbb{Z} \;\; \Longleftrightarrow \\
&\;\; \Longleftrightarrow \;\; x = \frac{\pi}{2} k, \; k\in \mathbb{Z}. \;\; \Longrightarrow\;\; \left \{\left (\frac{\pi}{2} k, 0\right )\,: \, k\in \mathbb{Z}\right \}.
\end{align*}

(II) Symmetry

None.

(III) Asymptotic Behavior

No vertical nor slant asymptotes.

Horizontal asymptotes.

\(\lim_{x\to \infty} e^{x/3}\sin 2x \; \text{ DNE}\;\;\; \text{ and } \;\;\; \lim_{x\to -\infty} e^{x/3}\sin 2x = 0 \)
for the second limit, apply the Squeeze Theorem to
\(-e^{x/3} \leq e^{x/3}\sin 2x \leq e^{x/3}. \)
Thus, the graph has horizontal asymptote \( y = 0\).

(IV) Monotonicity and Local Extrema

Critical numbers:
\begin{align*}
f'(x) =0 &\;\; \Longleftrightarrow \;\; \frac{1}{3}e^{x/3}\left (\sin 2x + 6 \cos 2x\right ) =0 \;\; \Longleftrightarrow \;\; \sin 2x + 6 \cos 2x =0 \;\; \Longleftrightarrow \\
& \;\; \underset{(*)}{\Longleftrightarrow} \;\; \sqrt{37}\sin\left (2x + \theta\right ) = 0 \;\; \Longleftrightarrow \;\; 2x + \theta = \pi k, \; k\in \mathbb{Z} \;\; \Longleftrightarrow x=-\frac{\theta}{2}+ \frac{\pi}{2} k, \; k\in \mathbb{Z}.
\end{align*}
Here, we can choose \(\theta\) as follows:
\(\left (\cos \theta , \sin \theta \right )= \left (\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right ) \Longrightarrow
\left [\renewcommand{\arraystretch}{2}\begin{array}{l}
\displaystyle \theta = \arccos \frac{1}{\sqrt{37}}\\
\displaystyle \theta = \arcsin \frac{6}{\sqrt{37}}
\end{array}\right . \)

Note that since \(\displaystyle \left (\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}\right )\) is in the first quadrant, both expressions for \(\theta\) are equivalent. It follows that the critical numbers are give by either one of the following two options:

\(
\left [\renewcommand{\arraystretch}{2}\begin{array}{l}
\displaystyle x = -\frac{1}{2}\arccos \frac{1}{\sqrt{37}}+ \frac{\pi}{2} k, \; k\in \mathbb{Z}\\
\displaystyle x = -\frac{1}{2}\arcsin \frac{6}{\sqrt{37}}+ \frac{\pi}{2} k, \; k\in \mathbb{Z}
\end{array}\right .
\)

To determine the intervals where \(f\) is increasing and where \(f\) is decreasing is now immediate as the graph of \(y=\sin\left (2x + \theta\right )\) is obtained by shifting the graph of \(y=\sin 2x\) to the left \(\theta\) units. Thus

\begin{align*}
f'(x) > 0 &\Longleftrightarrow \sin\left (2x + \theta\right ) > 0 \Longleftrightarrow 2\pi k < 2x + \theta < \pi + 2\pi k, \; k \in \mathbb{Z} \Longleftrightarrow \\
&\Longleftrightarrow -\frac{1}{2}\theta +\pi k < x < -\frac{1}{2}\theta+ \frac{1}{2}\pi + \pi k, \; k \in \mathbb{Z}
\end{align*}
\begin{align*}
f'(x) < 0 &\Longleftrightarrow \sin\left (2x + \theta\right ) < 0 \Longleftrightarrow \pi + 2\pi k < 2x + \theta < 2\pi + 2\pi k, \; k \in \mathbb{Z} \Longleftrightarrow \\
&\Longleftrightarrow -\frac{1}{2}\theta +\frac{1}{2}\pi +\pi k < x < -\frac{1}{2}\theta+\pi + \frac{1}{2}\pi + \pi k, \; k \in \mathbb{Z}
\end{align*}

It follows that each critical number of \(f\) yields either a local maximum value or a local minimum value (see Figure~\ref{ch07-07-f09}) and, in fact, we can explicitly identify the critical numbers that yield local maxima and those that yield local minima as follows:

local maxima at \(\displaystyle x= -\frac{1}{2}\arccos \frac{1}{\sqrt{37}} +\frac{1}{2}\pi +\pi k, \; k \in \mathbb{Z}\) (at these points \(f\) starts decreasing);

local minima at \(\displaystyle x= -\frac{1}{2}\arccos \frac{1}{\sqrt{37}} +\pi k, \; k \in \mathbb{Z}\) (at these points \(f\) starts increasing);

(V) Concavity and Inflection Points

We follow a similar procedure to the one used in (IV).

\begin{align*}
f”(x) =0 &\;\; \Longleftrightarrow \;\; \frac{1}{9}e^{x/3}\left (-35\sin 2x + 12 \cos 2x\right ) =0 \;\; \Longleftrightarrow \;\; -35\sin 2x + 12 \cos 2x =0 \;\; \Longleftrightarrow \\
& \;\; \underset{(*)}{\Longleftrightarrow} \;\; 37 \sin\left (2x + \phi\right ) = 0 \;\; \Longleftrightarrow \;\; 2x + \phi = \pi k, \; k\in \mathbb{Z} \;\; \Longleftrightarrow x=-\frac{\phi}{2}+ \frac{\pi}{2} k, \; k\in \mathbb{Z}.
\end{align*}
Here, \( \phi \) satisfies \(\displaystyle \left (\cos \phi , \sin \phi \right )= \left (-\frac{35}{37},\frac{12}{37}\right )\). Note that this point is in the second quadrant, thus
\(
\left [\renewcommand{\arraystretch}{2}\begin{array}{l}
\displaystyle \phi = \arccos\left (-\frac{35}{37} \right )\\
\displaystyle \phi = \pi-\arcsin \frac{12}{37}
\end{array}\right .. \)

The first option is immediate because the range of \(y=\arccos x\) is \([0, \pi]\) (first and second quadrants), but the range for \(y = \arcsin x\) is \(\displaystyle \left [-\frac{\pi}{2}, \frac{\pi}{2}\right ]\) (first and fourth quadrants). Thus, \(\displaystyle \arcsin \frac{12}{37}\) is in the first quadrant, and it needs to be adjusted to be in the second quadrant (use the unit circle). It follows that
\(f”(x) = 0 \Longleftrightarrow
\left [\renewcommand{\arraystretch}{2}\begin{array}{l}
\displaystyle x = -\frac{1}{2} \arccos\left (-\frac{35}{37} \right )+ \frac{\pi}{2} k, \; k\in \mathbb{Z}\\
\displaystyle x = -\frac{1}{2}\left (\pi-\arcsin \frac{12}{37} \right )+ \frac{\pi}{2} k, \; k\in \mathbb{Z}
\end{array}\right .
\)

These expressions for the zeroes of the second derivative are equivalent and, as we shall see, yield the inflection points of \(f\).

To determine the intervals where \(f\) is concave upward and where it is concave downward, we proceed as before (in (IV)). Note that the graph of \(y = \sin\left (2x + \phi\right )\) is obtained by shifting the graph of \(y = \sin 2x\) to the left \(\phi\) units.

\begin{align*}
f”(x) > 0 &\Longleftrightarrow \sin\left (2x + \phi\right ) > 0 \Longleftrightarrow 2\pi k < 2x + \phi < \pi + 2\pi k, \; k \in \mathbb{Z} \Longleftrightarrow \\
&\Longleftrightarrow -\frac{1}{2}\phi +\pi k < x < -\frac{1}{2}\phi+ \frac{1}{2}\pi + \pi k, \; k \in \mathbb{Z}
\end{align*}
\begin{align*}
f”(x) < 0 &\Longleftrightarrow \sin\left (2x + \phi\right ) < 0 \Longleftrightarrow \pi + 2\pi k < 2x + \phi < 2\pi + 2\pi k, \; k \in \mathbb{Z} \Longleftrightarrow \\
&\Longleftrightarrow -\frac{1}{2}\phi +\frac{1}{2}\pi +\pi k < x < -\frac{1}{2}\phi+\pi + \frac{1}{2}\pi + \pi k, \; k \in \mathbb{Z}
\end{align*}

It follows that each zero of \(f”\) yields an inflection point. See Figure~\ref{ch07-07-f09}.

(VI) Graph

Based on all the above information, the graph is as shown in Figure~\ref{ch07-07-f09}.

Figure 7.40 Example 2

Note, as pointed out above, that
\(-e^{x/3} \leq e^{x/3}\sin 2x \leq e^{x/3}, \)
as the graph clearly shows.

7.7.4 Logarithmic Functions

Example 1 Given the function \(f\) together with its first and second derivatives,
\[f(x) =\frac{\ln x^2}{x^2},\;\;\; f'(x) = -\frac{2\left (\ln x^2 – 1\right )}{x^3}, \;\;\; f”(x) = \frac{2\left (3\ln x^2 – 5\right )}{x^4}, \]
sketch its graph by completing steps (I)–(VI).

(I) Domain and Intercepts

\(D_f=(-\infty, 0)\cup(0,\infty)\). \(x\)-intercepts: \((\pm 1,0)\); \(y\)-intercept: none.

(II) Symmetry

The function is even (explain why). Thus, the graph is symmetry with respect to the \(y\)-axis.

(III) Asymptotic Behavior

Vertical Asymptotes

We have
\(\lim_{x\to 0} \ln x^2 = -\infty. \;\; \Longrightarrow \;\; \lim_{x\to 0} \frac{\ln x^2}{x^2}= -\infty, \;\; \Longrightarrow \;\; \text{ vertical asymptote: } x=0. \)

Horizontal Asymptotes

\(
\lim_{x\to \infty} \frac{\ln x^2}{x^2} = \lim_{x\to \infty} \frac{2\ln x}{x^2}
\;\;\underset{\text{l’H}}{=} \;\;
\lim_{x\to \infty} \frac{2}{2x^2} = 0, \)

and, because of symmetry, \(\displaystyle \lim_{x\to – \infty} \frac{\ln x^2}{x^2} = 0 \). Thus \( y = 0\) is the only horizontal asymptote.

(IV) Monotonicity and Local Extrema

Since
\(f'(x) = -\frac{2\left (\ln x^2 – 1\right )}{x^3}= 0 \;\;\Longleftrightarrow \;\; \ln x^2 – 1=0 \;\;\Longleftrightarrow \;\; x^2 = e \;\;\Longleftrightarrow \;\; x =\pm \sqrt{e}, \)
\(f\) has critical numbers: \(\displaystyle \left \{\pm \sqrt{e}\right \}\).

To determine the sign of \(f’\), we are going to restrict our attention to the case where \(x > 0 \). Then
\(f'(x) = -\frac{2\left (\ln x^2 – 1\right )}{x^3} = -\frac{2\left (2\ln x – 1\right )}{x^3} > 0 \;\;\Longleftrightarrow\;\; 2\ln x – 1 < 0 \;\;\Longleftrightarrow\;\; 0 < x < \sqrt{e}. \)

\(f\) is increasing on \(\left (0, \sqrt{e}\right )\), and decreasing on \(\left (\sqrt{e}, \infty\right )\).

It follows that \(f\) has a local maxim value \(\displaystyle f\left (\sqrt{e}\right )=\frac{1}{e}\) and no local maximum values on \((0, \infty)\).

(V) Concavity and Inflection Points

Since \(\displaystyle f”(x) = \frac{2\left (3\ln x^2 – 5\right )}{x^4}\)
\(f”(x) = 0 \;\;\Longleftrightarrow\;\; 3\ln x^2 – 5 = 0 \;\;\Longleftrightarrow\;\; x^2 = e^{5/3}\;\;\Longleftrightarrow\;\; x = \pm e^{5/6}. \)

Proceeding in a similar fashion as in (IV), for \(x > 0 \) we have

\(
f”(x) = \frac{2\left (3\ln x^2 – 5\right )}{x^4} = \frac{2\left (6\ln x – 5\right )}{x^4} > 0 \;\;\Longleftrightarrow\;\; 6\ln x – 5 > 0 \;\;\Longleftrightarrow\;\; x > e^{5/6}. \)

Thus, on the interval \((0, \infty)\), \(f\) is concave upward on \(\displaystyle \left ( e^{5/6}, \infty\right )\), and downward on \(\displaystyle \left (0, e^{5/6} \right )\).

Hence \(f\) has an inflection point at \(x = e^{5/6}\).

(VI) Graph

Based on all the above information (recall that \(f\) is even), the graph is as shown in Figure Figure 7.42

Figure 7.42: Example 1