Answer
(a) 25
(b) $f'(10) \lt f'(30)$
(c) $f'(60) \gt \frac{f(80)-f(40)}{80-40}$
Work Step by Step
(a) On the graph, we can see that $f'(50)$ (which is the slope at x=50) is approximately equal to the average rate of change on the interval $[40,60]$.
$f'(50) \approx \frac{700-200}{60-40} = 25$
(b) On the graph, we can see that $f'(10)$ (which is the slope at x = 10) is more negative than $f'(30)$ (which is the slope at x = 30).
$f'(10) \lt f'(30)$
(c) $\frac{f(80)-f(40)}{80-40}$ is the slope of a straight line which we could draw from the point $(40, f(40))$ to the point $(80, f(80))$. On the graph, we can see that this line would have a slope which is less than the slope at $x = 60$.
Therefore:
$f'(60) \gt \frac{f(80)-f(40)}{80-40}$
