
Originally Posted by
Soroban
Hello, addanis!
Set the derivative equal to 0 and solve.
. . $\displaystyle g'(x) \:=\:\cos x + \sin x \:=\:0\quad\Rightarrow\quad \sin x \:=\:-\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x}\:=\:-1$
. . $\displaystyle \tan x \:=\:-1\quad\Rightarrow\quad x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}$
Second Derivative Test: .$\displaystyle g''(x) \:=\:-\sin(x) + \cos(x)$
$\displaystyle g''\left(\frac{3\pi}{4}\right) \:=\:-\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) \:=\:-\sqrt{2}$ . . . negative: concave down $\displaystyle \cap$
. . . . Hence: relative maximum at $\displaystyle \left(\frac{3\pi}{4},\:\sqrt{2}\right)$
$\displaystyle g''\left(\frac{7\pi}{4}\right)\:=\:-\sin\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) \:=\:+\sqrt{2}$ . . . positive: concave up $\displaystyle \cup$
. . . . Hence, relative minimum at $\displaystyle \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right)$
Test endpoints:
. . $\displaystyle g(0)\:=\:\sin(0) - \cos(0) \:=\:-1$
. . $\displaystyle g(\pi)\:=\:\sin(\pi) - \cos(\pi) \:=\:+1$
Therefore . . . .$\displaystyle \begin{array}{cc}\text{Abs. maximum: }& \left(\frac{3\pi}{4},\:\sqrt{2}\right) \\ \text{Abs. minimum: }& \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right) \end{array} $