Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 32

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = x + \cos x$

The guidelines of Curve Sketching
A. Domain.
We know that the function is continuous everywhere. Thus, the domain is $(-\infty, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x=0$
$y = 0 + \cos 0 = 0 + 1 = 1$
Solving for $x$-intercept, when $y=0$
$0 = x + \cos x$
By trial and error,
$x \approx - 0.75$

C. Symmetry.
The function is not symmetric to either $y$-axis or origin by using symmetry test.


D. Asymptotes.
Since the domain is $(-\infty,\infty)$. The function has no vertical asymptotes. Also, since $\displaystyle \lim_{x \to \infty} f(x) = \infty \text{ and } \lim_{x \to \infty} f(x) = -\infty$, the function has no horizontal asymptotes.

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$,
$f'(x) = 1 - \sin x$
Since $f'(x) > 0$ for all $x$, It means that the function is always increasing on the interval $(-\infty, \infty)$

F. Local Maximum and Minimum Values.
Since $f'(x)$ is always greater than 0, the function has no local maxima and minima.


G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= 1 - \sin x \text{, then}\\
\\
f''(x) &= -\cos x\\
\\
\\
\text{when } f''(x) &= 0 \\
\\
0 &= -\cos x
\end{aligned}
\end{equation}
$


Therefore, $\displaystyle x = \frac{\pi}{2} + 2 \pi n \text{ and } x = \frac{3\pi}{2} + 2\pi n$ where $n$ is any integer.
Suppose that the function is defined only for interval $[-2\pi, 2 \pi]$
Hence, the concavity is...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < \frac{-3\pi}{2} & - & \text{Downward}\\
\hline\\
\frac{-3\pi}{2} < x < \frac{-\pi}{2} & + & \text{Upward}\\
\hline\\
\frac{-\pi}{2} < x < \frac{\pi}{2} & - & \text{Downward}\\
\hline\\
\frac{\pi}{2} < x < \frac{3\pi}{2} & + & \text{Upward}\\
\hline\\
x > \frac{3\pi}{2} & - & \text{Downward}\\
\hline
\end{array}
$


Therefore, we have inflection points at interval $[-2 \pi, 2\pi]$,

$
\begin{equation}
\begin{aligned}
f \left( \frac{\pi}{2} \right) &= \frac{\pi}{2} + \cos \frac{\pi}{2} = \frac{\pi}{2}\\
\\
f \left( \frac{-3\pi}{2} \right) &= \frac{-3\pi}{2}\\
\\
f \left( \frac{3\pi}{2} \right) &= \frac{3\pi}{2}\\
\\
f \left( \frac{-\pi}{2} \right) &= \frac{-\pi}{2}
\end{aligned}
\end{equation}
$


H. Sketch the Graph.

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