Question

Let f be a differentiable function with f(0)=-4 and f(10)=11. which of the following must be true for some c in the interval (0,10)?

A) f'(c)=0, since the Extreme Value Theorem applies

B) f'(c)=11+(-4)/10-0 since the Mean Value Theorem applies

C) f'(c)=11-(-4)/10-0, since the Mean Value Theorem applies

D) f'(c)=1.5, since the Intermediate Value Theorem applies

Answer 1

The mean value theorem states that if $f$ is a continuous function on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ , then there exists a point $c$ in $(a,b)$ such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

We have:

$\frac{f(b)-f(a)}{b-a}=\frac{11-(-4)}{10-0}=f'(c)$ , $c\isin (0,10)$

Answer: C) f'(c)=11-(-4)/10-0, since the Mean Value Theorem applies