$x=0$  

 $x=-4$  

 $x=-2$  

 $x=4$  

a local maximum at   $x=-2$  

a local maximum at  $x=4$  

local maxima at  $x=-2$  and  $x=4$  

no local extrema

$1$ because there is only one critical point

$2$ because parabolas increase then decrease (or vice versa)

Depends on the parabola

 $f\left(x\right)$  is not defined at  $x=1$  

 $1$  

 $\frac{2}{\sqrt{3}}$  

 $f'\left(1\right)\$  does not exist

 $\frac{1}{5}$  

 $-\frac{1}{5}$  

 $\frac{\sqrt{34}}{3}$  

 $-\frac{\sqrt{34}}{3}$  

$f\left(x\right)\ =\ \left(x-2\right)^{\frac{2}{3}}\$ on $\left[3,5\right]$

$f\left(x\right)\ =\ \sqrt{1-x^2}$ on $\left[-1,1\right]$

$f\left(x\right)=\left(2x-3\right)^4$ on $\left[1,2\right]$

$f\left(x\right)=e^{-x^2}$ on $\left[-4,4\right]$

positive

negative

zero

undefined

absolute maximum

absolute minimum

relative maximum

relative minimum

I should cut out squares that are 1/2 in by 1/2 in

I should cut out squares that are 1 in by 1 in

I should cut out squares that are 3 in by 3 in

I should not cut out any squares

endpoint

zero

maximum

undefined

undefined

positive

negative

zero