Single Variable Calculus Questions and Answers

Start Your Free Trial

Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 40

Expert Answers info

eNotes eNotes educator | Certified Educator

calendarEducator since 2007

write13,548 answers

starTop subjects are Math, Literature, and Science

Suppose that $f(x) = x + \cos x, - 2 \pi \leq x \leq 2 \pi$

a.) Determine the intervals of increase or decrease.

If $f(x) = x + \cos x$ then,

$ \begin{equation} \begin{aligned} f'(x) =& 1 - \sin x \\ \\ f''(x) =& - \cos x \end{aligned} \end{equation} $

To find the critical numbers, we set $f'(x) = 0$, so..

$ \begin{equation} \begin{aligned} f'(x) = 0 =& 1 - \sin x \\ \\ 0 =& 1 - \sin x \\ \\ \sin x =& 1 \\ \\ x =& \sin^{-1} [1] \\ \\ x =& \frac{\pi}{2} + 2 \pi n \text{ where $n$ is an integer} \end{aligned} \end{equation} $

For the interval $-2 \pi \leq x \leq 2 \pi$, the critical numbers are

$\qquad \displaystyle x = \frac{\pi}{2}, x = - \frac{3}{2} \pi$

Hence, we can divide the interval by:

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f \\ \hline\\ \displaystyle -2 \pi < x - \frac{3 \pi}{2} & + & \text{increasing on} \left( -2 \pi, - \frac{3 \pi}{2} \right) \\ \hline\\ \displaystyle - \frac{3 \pi}{2} < x < \frac{\pi}{2} & + & \text{increasing on} \displaystyle \left( \frac{-3 \pi}{2}, \frac{\pi}{2} \right) \\ \hline\\ \displaystyle \frac{\pi}{2} < x < 2 \pi & + & \text{increasing on} \left( \frac{\pi}{2}, 2 \pi \right)\\ \hline \end{array} $

These data obtained by substituting any values of $x$ to $f'(x)$ within the specified interval. Check its sign, if it's positive, it means that the curve is increasing on that interval. On the other hand, if the sign is negative, it means that the curve is decreasing on that interval.

b.) Find the local maximum and minimum values.

We will use Second Derivative Test to evaluate $f''(x)$ at these critical numbers:

So when $\displaystyle x = \frac{\pi}{2},$

$ \begin{equation} \begin{aligned} f'' \left( \frac{\pi}{2} \right) =& - \cos \left( \frac{\pi}{2} \right) \\ \\ f''\left( \frac{\pi}{2} \right) =& 0 \end{aligned} \end{equation} $

Since $\displaystyle f''\left( \frac{\pi}{2} \right)$ and $\displaystyle f'' \left( \frac{- 3 \pi}{2} \right) = 0$ the function has no local maximum and minimum.

c.) Find the intervals of concavity and the inflection points.

We set $f''(x) = 0$, to determine the inflection points..

$ \begin{equation} \begin{aligned} f''(x) = 0 =& - \cos x \\ \\ \cos x =& 0 \\ \\ x =& \cos ^{-1} [0] \\ \\ x =& \pm \frac{\pi}{2} + 2 \pi n' \text{ where $n$ is an integer} \end{aligned} \end{equation} $

For the interval $-2 \pi \leq x \leq 2 \pi$, the inflection points are

$\qquad \displaystyle x = \frac{\pi}{2}, x = \frac{-3 \pi}{2}, x = \frac{- \pi}{2}, x = \frac{3 \pi}{2}$

Let's divide the interval to determine the concavity..

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

These values are obtained by evaluating $f''(x)$ within the specified interval. The concavity is upward when the sign of $f''(x)$ is positive. On the other hand, the concavity is downward when the sign of $f''(x)$ is negative.

d.) Using the values obtained, illustrate the graph of $f$.