Find the extrema of the function on the given interval, and say where they occur.
-$\sin x + \cos x , 0 \leq x \leq 2 \pi$
A) local maxima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 5 \pi } { 4 }$;
local minima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$

B) local maxima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$;
local minima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 5 \pi } { 4 }$

C) local maxima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$;
local minima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 7 \pi } { 4 }$

D) local maxima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 7 \pi } { 4 }$;
local minima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$

Question

Find the extrema of the function on the given interval, and say where they occur.
-$\sin x + \cos x , 0 \leq x \leq 2 \pi$
A) local maxima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 5 \pi } { 4 }$;
local minima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$

B) local maxima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$;
local minima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 5 \pi } { 4 }$

C) local maxima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$;
local minima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 7 \pi } { 4 }$

D) local maxima: 1 at $x = 0$ and $- \sqrt { 2 }$ at $x = \frac { 7 \pi } { 4 }$;
local minima: 1 at $x = 2 \pi$ and $\sqrt { 2 }$ at $x = \frac { \pi } { 4 }$

Quizplus · Accepted Answer

To find the extrema, we take the derivative of the function and set it equal to zero. The derivative of $\sin x + \cos x$ is $\cos x - \sin x$. Setting this equal to zero gives $\cos x = \sin x$ which implies $	an x = 1$ or $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. Checking the values of the function at these critical points, we find that $\sqrt{2}$ and $- \sqrt{2}$ are local maxima and $1$ is a local minimum at $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ respectively. Additionally, we find that $1$ is a local maximum and $- \sqrt{2}$ is a local minimum at $0$ and $2\pi$ respectively by checking the endpoints of the interval. Thus, the extrema occur at $x = 0, \frac{\pi}{4}, 2\pi, \frac{5\pi}{4}$ and the corresponding function values are $1, \sqrt{2}, 1, -\sqrt{2}$ respectively. The correct answer is choice B.

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