Find the points of inflection and discuss the concavity of the function $f ( x ) = - 8 x - 2 \cos x$ on the interval $[ 0,2 \pi ]$
A) concave down on $( 0,2 \pi )$; no points of inflection
B) concave downward on $\left( \frac { \pi } { 2 } , \frac { 3 \pi } { 2 } \right) ;$ concave upward on $\left( 0 , \frac { \pi } { 2 } \right) , \left( \frac { 3 \pi } { 2 } , 2 \pi \right)$; inflection points at $x = \frac { \pi } { 2 }$ and $x = \frac { 3 \pi } { 2 }$
C) concave upward on $\left( \frac { \pi } { 2 } , \frac { 3 \pi } { 2 } \right) ;$ concave downward on $\left( 0 , \frac { \pi } { 2 } \right) , \left( \frac { 3 \pi } { 2 } , 2 \pi \right)$ inflection points at $x = \frac { \pi } { 2 }$ and $x = \frac { 3 \pi } { 2 }$
D) concave up on $( 0,2 \pi )$; no points of inflection
E) none of the above

Question

Quizplus · Accepted Answer

To find the concavity of the function, we need to find the second derivative of the function:

$$f'(x) = -8 + 2\sin x$$

$$f''(x) = 2\cos x $$

Setting $f''(x) = 0$ gives us inflection points:

$$2\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$$

We also need to check the concavity of the function on different intervals:

On $\left(0,\frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right)$, since $f''(x) > 0$ (cosine is positive), the function is concave upward.

On $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$, since $f''(x) < 0$ (cosine is negative), the function is concave downward.

Therefore, the best choice is B: concave downward on $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$; concave upward on $\left(0,\frac{\pi}{2}\right), \left(\frac{3\pi}{2},2\pi\right)$; inflection points at $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$.

Find the Points of Inflection and Discuss the Concavity of the Function