Answer
(a) $f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,8]$.
(b) $c=2~~$ or $~~c=6.2$
(approximate values)
(c) $c=3.5~~$ or $~~c = 5.5$
(approximate values)
Work Step by Step
(a) 1. We can see that $f$ is continuous on the closed interval $[0,8]$
2. $f$ is a smooth curve with no sharp points so $f$ is differentiable on the open interval $(0,8)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,8]$.
(b) $\frac{f(8) -f(0)}{8-0} = \frac{4-1}{8} = \frac{3}{8}$
According to the Mean Value Theorem, there is a $c$ in the interval $(0,8)$ such that $f'(c) = \frac{3}{8}$
On the graph, we can see that there are two points with this slope.
When $c = 2$, then $f'(c) \approx \frac{3}{8}$
When $c = 6.2$, then $f'(c) \approx \frac{3}{8}$
(c) $\frac{f(6) -f(2)}{6-2} = \frac{1-3}{4} = -\frac{1}{2}$
According to the Mean Value Theorem, there is a $c$ in the interval $(2,6)$ such that $f'(c) = -\frac{1}{2}$
On the graph, we can see that there are two points with this slope.
When $c = 3.5$, then $f'(c) \approx -\frac{1}{2}$
When $c = 5.5$, then $f'(c) \approx -\frac{1}{2}$
