Answer
a) Critical point c is 2. $f(c)=-3$
b) $f(0)=1, f(4)=1$
c) $1$ is the maximum value and $-3$ is the minimum value.
d) Maximum value$=1$, minimum value $=-2$
Work Step by Step
a) $f'(x)=2x-4=0\implies 2x=4\implies x=\frac{4}{2}=2$
The critical point is $x=2$.
$f(c)=(2)^{2}-4(2)+1=-3$
b) $f(0)=(0)^{2}-4(0)+1=1$
$f(4)=(4)^{2}-4(4)+1=1$
c) Comparing the values of $f$ on the end points and the critical point, we find that
$f(4)=f(0)=1$ is the maximum value and
$f(2)=-3$ is the minimum value.
d) The critical point of $f$ is outside the interval. So the extreme values are obtained by comparing the values of $f$ on the end points. We find that
$f(0)=1$ is the maximum value and
$f(1)=(1)^{2}-4(1)+1=-2$ is the minimum value.
