Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 34

Use the guidelines of curve sketching to sketch the curve $\displaystyle y = 2x - \tan x, - \frac{\pi}{2} < x < \frac{\pi}{2}$ then find the equation of slant asymptote.

A. Domain. $\displaystyle \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) $

B. Intercepts. Solving for $y$ intercept, when $x = 0$,

$\displaystyle y = 2(0) - \tan(0) = 0$

Solving for $x$ intercept, when $y = 0$,


$
\begin{equation}
\begin{aligned}

0 =& 2x - \tan x

\end{aligned}
\end{equation}
$


By trial and error, we have

$x = 1.15$ and $x = -1.15$

C. Symmetry. Since $f(-x) = -f(x)$, the function is symmetric to the origin.

D. Asymptote. The function has no asymptotes.

E. Intervals of increase or decrease,

If we take two derivative of $f(x)$

$f'(x) = 2 - \sec^2 x$

when $f'(x) = 0$


$
\begin{equation}
\begin{aligned}

0 =& 2 - \sec^2 x
\\
\\
\sec^2 x =& 2
\\
\\
\cos^2 x =& \frac{1}{2}
\\
\\
x =& \cos^{-1} \left[ \sqrt{\frac{1}{2}} \right]
\\
\\
x =& \pm \frac{\pi}{4} + 2 \pi n

\end{aligned}
\end{equation}
$


For the interval $\displaystyle - \frac{\pi}{2} < x < \frac{\pi}{2}$,

The critical numbers are,

$\displaystyle x = \frac{\pi}{4}$ and $\displaystyle x = - \frac{\pi}{4}$

Hence, the intervals of increase and decrease are..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
\displaystyle - \frac{\pi}{2} < x < - \frac{\pi}{4} & - & \displaystyle \text{decreasing on } \left(- \frac{\pi}{2}, \frac{\pi}{4} \right) \\
\displaystyle - \frac{\pi}{4} < x < \frac{\pi}{7} & + & \displaystyle \text{increasing on } \left(- \frac{\pi}{4}, \frac{\pi}{4} \right) \\
\displaystyle \frac{\pi}{4} < x < \frac{\pi}{2} & - & \displaystyle \text{decreasing on } \left(\frac{\pi}{4}, \frac{\pi}{2} \right) \\
\hline

\end{array}
$


F. Local Maximum and Minimum Values

Since $f'(x)$ changes from negative to positive at $\displaystyle x = - \frac{\pi}{4}, f\left( - \frac{\pi}{4} \right) = -0.5708$ is a local minimum. On the other hand, since $f'(x)$ changes from negative to positive at $\displaystyle x = \frac{\pi}{4}, f\left( \frac{\pi}{4} \right) = 0.5708$ is a local maximum.

G. Concavity and inflection point

If $f'(x) = 2 - \sec^2x$, then


$
\begin{equation}
\begin{aligned}

f''(x) =& -2 \sec x (\sec x \tan x) = -2 \sec ^ 2 x \tan x
\\
\\
f''(x) =& \frac{-2 \sin x}{ \cos ^ 3 x}

\end{aligned}
\end{equation}
$


when $f''(x) = 0,$

$\displaystyle 0 = - 2 \sin x$

We have,

$x = 0$

Therefore, the inflection point is at

$f(0) = 0$

Hence, the concavity is..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
x < 0 & + & \text{Upward} \\
x > 0 & - & \text{Downward}\\
\hline

\end{array}
$

H. Sketch the graph.