Find the extrema of the function on the given interval, and say where they occur.
-$\sin 4 x , 0 \leq x \leq \frac { \pi } { 2 }$
A) local maxima: 1 at $x = \frac { \pi } { 8 }$ and 0 at $x = \frac { \pi } { 2 }$;
local minima: 0 at $x = 0$ and $- 1$ at $x = \frac { 3 \pi } { 8 }$

B) local maxima: 1 at $x = \frac { \pi } { 8 }$ and 0 at $x = \frac { \pi } { 2 }$;
local minimum: $- 1$ at $x = \frac { 3 \pi } { 8 }$

C) local maxima: 1 at $x = \frac { \pi } { 4 }$ and 0 at $x = \frac { \pi } { 2 }$;
local minima: 0 at $x = 0$ and $- 1$ at $x = \frac { 3 \pi } { 8 }$

D) local maxima: 1 at $x = \frac { \pi } { 8 }$ and 0 at $x = \frac { \pi } { 4 }$;

local minimum: 0 at $x = 0$

Question

Find the extrema of the function on the given interval, and say where they occur.
-$\sin 4 x , 0 \leq x \leq \frac { \pi } { 2 }$
A) local maxima: 1 at $x = \frac { \pi } { 8 }$ and 0 at $x = \frac { \pi } { 2 }$;
local minima: 0 at $x = 0$ and $- 1$ at $x = \frac { 3 \pi } { 8 }$

B) local maxima: 1 at $x = \frac { \pi } { 8 }$ and 0 at $x = \frac { \pi } { 2 }$;
local minimum: $- 1$ at $x = \frac { 3 \pi } { 8 }$

C) local maxima: 1 at $x = \frac { \pi } { 4 }$ and 0 at $x = \frac { \pi } { 2 }$;
local minima: 0 at $x = 0$ and $- 1$ at $x = \frac { 3 \pi } { 8 }$

D) local maxima: 1 at $x = \frac { \pi } { 8 }$ and 0 at $x = \frac { \pi } { 4 }$;

local minimum: 0 at $x = 0$

Quizplus · Accepted Answer

The function $\sin 4x$ has a period of $\frac{\pi}{2}$, so within the given interval $0 \leq x \leq \frac{\pi}{2}$, it completes one full cycle. The maximum value of $\sin$ is 1, which occurs at $x = \frac{\pi}{8}$ (since $\sin 4(\frac{\pi}{8}) = \sin(\frac{\pi}{2}) = 1$), and the minimum value is -1, which occurs at $x = \frac{3\pi}{8}$ (since $\sin 4(\frac{3\pi}{8}) = \sin(3\frac{\pi}{2}) = -1$). At $x = 0$ and $x = \frac{\pi}{2}$, the function values are 0, but these are not local extrema within the cycle of the sine function; they represent the starting and ending points of one complete cycle on the given interval.

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