#### Answer

a) Minimum: (1, -1), Maximum: (-1, 3)
b) Maximum: (3, 3)
c) Minimum: (1, -1)
d) Minimum: (1, -1)

#### Work Step by Step

Given: $f(x)=x^{2}-2x$
Find critical number(s).
$f'(x)=2x-2$ (take the derivative of $f(x)$)
$f'(x)=2(x-1) = 0$ (x is a critical number when $f'(x)=0$)
$0 = x-1$ (divide both sides by 2, then add 1 to both sides)
$x=1$ (there is a critical number when $x=1$)
a) Given: $[-1, 2]$
Since the interval is closed, plug in both -1, 2 AND the critical number 1 to the function $f(x)$
$f(-1)=(-1)^{2}-2(-1)=3$
$f(1)=1^{2}-2(1)=-1$
$f(2)=2^{2}-2(2)=0$
The y value was smallest at $(1, -1)$ and biggest at $(-1, 3)$,
so absolute minimum at $(1, -1)$ and absolute maximum at $(-1, 3)$
b) Given: $(1,3]$
The function is open at $x=1$ so we don't plug in $x=1$ to the function, even though it's a critical number. We still have to plug in the x value when the function is closed, when $x=3$
$f(3)=3^{2}-2(3)=3$
So absolute maximum at $(3, 3)$
c) Given: $(0,2)$
The interval is open interval, just plug in the critical number 1 to the function.
$f(x)=1^{2}-2(1)=-1$
So absolute minimum at $(1, -1)$
d) [1,4)
The interval is open at $x=4$ so we don't plug in $x=4$ to the function, we just need to plug in the critical number/closed interval 1 to the function.
$f(x)=1^{2}-2(1)=-1$
So absolute minimum at $(1, -1)$