1. ## [SOLVED] Calculus 1

Find the absolute maximum and absolute minimum of g(x)= sin(x) - cos(x) on [0,π].

It is on a closed interval on zero and pie.

Thanks.

Find the absolute maximum and absolute minimum of g(x)= sin(x) - cos(x) on [0,π].

It is on a closed interval on zero and pie.

Thanks.
$f(x)=\sin x-\cos x$.
This is continous on $[0,\pi]$.
By Fermat's principle a relative extrema occurs when the derivative does not exist or is zero. Since the derivative is zero the only points on the interval $(0,\pi)$ is when derivative is zero. And we need to check the boundardy points because they are not gaurennted by the theorem.
$g'(x)=\cos x+\sin x$
Thus,
$\cos x+\sin x=0$
$\cos x=-\sin x$, $\cos x\not = 0$
$1=-\tan x$
$\tan x=-1$
$x=\pi -\frac{\pi}{4}=\frac{3\pi}{4}$
Und,
$g(3\pi/4)=\sqrt{2}/2+\sqrt{2}/2=\sqrt{2}$
$g(0)=0-1=-1$
$g(\pi)=0+1=1$
Thus,
$x=3\pi/4$---> Max
$x=0$---> Min

Find the absolute maximum and minimum of $g(x) \:= \:\sin(x) - \cos(x)$ on $[0,\,\pi]$

Set the derivative equal to 0 and solve.

. . $g'(x) \:=\:\cos x + \sin x \:=\:0\quad\Rightarrow\quad \sin x \:=\:-\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x}\:=\:-1$

. . $\tan x \:=\:-1\quad\Rightarrow\quad x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}$

Second Derivative Test: . $g''(x) \:=\:-\sin(x) + \cos(x)$

$g''\left(\frac{3\pi}{4}\right) \:=\:-\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) \:=\:-\sqrt{2}$ . . . negative: concave down $\cap$
. . . . Hence: relative maximum at $\left(\frac{3\pi}{4},\:\sqrt{2}\right)$

$g''\left(\frac{7\pi}{4}\right)\:=\:-\sin\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) \:=\:+\sqrt{2}$ . . . positive: concave up $\cup$
. . . . Hence, relative minimum at $\left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right)$

Test endpoints:
. . $g(0)\:=\:\sin(0) - \cos(0) \:=\:-1$
. . $g(\pi)\:=\:\sin(\pi) - \cos(\pi) \:=\:+1$

Therefore . . . . $\begin{array}{cc}\text{Abs. maximum: }& \left(\frac{3\pi}{4},\:\sqrt{2}\right) \\ \text{Abs. minimum: }& \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right) \end{array}$

4. Originally Posted by Soroban

Set the derivative equal to 0 and solve.

. . $g'(x) \:=\:\cos x + \sin x \:=\:0\quad\Rightarrow\quad \sin x \:=\:-\cos x\quad\Rightarrow\quad \frac{\sin x}{\cos x}\:=\:-1$

. . $\tan x \:=\:-1\quad\Rightarrow\quad x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}$

Second Derivative Test: . $g''(x) \:=\:-\sin(x) + \cos(x)$

$g''\left(\frac{3\pi}{4}\right) \:=\:-\sin\left(\frac{3\pi}{4}\right) + \cos\left(\frac{3\pi}{4}\right) \:=\:-\sqrt{2}$ . . . negative: concave down $\cap$
. . . . Hence: relative maximum at $\left(\frac{3\pi}{4},\:\sqrt{2}\right)$

$g''\left(\frac{7\pi}{4}\right)\:=\:-\sin\left(\frac{7\pi}{4}\right) + \cos\left(\frac{7\pi}{4}\right) \:=\:+\sqrt{2}$ . . . positive: concave up $\cup$
. . . . Hence, relative minimum at $\left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right)$

Test endpoints:
. . $g(0)\:=\:\sin(0) - \cos(0) \:=\:-1$
. . $g(\pi)\:=\:\sin(\pi) - \cos(\pi) \:=\:+1$

Therefore . . . . $\begin{array}{cc}\text{Abs. maximum: }& \left(\frac{3\pi}{4},\:\sqrt{2}\right) \\ \text{Abs. minimum: }& \left(\frac{7\pi}{4},\:\text{-}\sqrt{2}\right) \end{array}$

But the interval is $[0,\pi]$