Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 36

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \sec x + \tan x \quad ,0 < x < \frac{\pi}{2}$

The guidelines of Curve Sketching
A. Domain.
The function has domain $\displaystyle \left( 0, \frac{\pi}{2} \right)$

B. Intercepts.
Solving for $y$-intercept, when $x=0$
$y = \sec 0 + \tan 0 = 1 + 0 = 1$

C. Symmetry.
By using symmetry test, we can say that the function is not symmetric to either $y$-axis or origin.

D. Asymptotes.
We can rewrite $\displaystyle y = \sec x + \tan x \text{ as } y = \frac{1}{\cos x} + \frac{\sin x}{\cos x} \quad y = \frac{1 + \sin x}{\cos x}$
when $\cos x = 0$,
$\displaystyle x = \frac{\pi}{2} + 2\pi n$, where $n$ is any integet
For interval $\displaystyle 0 < x < \frac{\pi }{2}$, the vertical asymptote is none.

E. Intervals of Increase or Decrease.

$
\begin{equation}
\begin{aligned}
\text{if } f(x) &= \sec x + \tan x, \text{ then}\\
\\
f'(x) &= \sec x \tan x + \sec^2 x\\
\\
f'(x) &= \frac{\sin x}{\cos ^2 x} + \frac{1}{\cos^2 x} = \frac{\sin x + 1}{\cos^2 x}\\
\\
\\
\text{when } f'(x) &= 0 \\
\\
0 &= \sin x + 1\\
\\
\sin x &= -1\\
\\
x &= \frac{-\pi}{2} + 2 \pi n \qquad \text{where } n \text{ is any integer}
\end{aligned}
\end{equation}
$

For interval $\displaystyle 0 < x < \frac{\pi }{2}$,
The function has no critical numbers.
Thus the interval of increase or decrease is...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
0 < x < \frac{\pi}{2} & + & \text{increasing on } (0,\frac{\pi}{2})\\
\hline
\end{array}
$


F. Local Maximum and Minimum Values.
Since $f'(x)$ is always increasing on interval $\displaystyle 0 < x < \frac{\pi }{2}$. It means that we have no local maxima and minima at this interval

G. Concavity and Points of Inflection.
Since $f(x)$ is always increasing, we can say that the function has no inflection points. Therefore, the function has no upward concavity at $\displaystyle \left( 0, \frac{\pi}{2} \right)$

H. Sketch the Graph.