Answer

Solution

1. 1

   By the Mean Value Theorem on $[1/2,1]$, there exists $c\in(1/2,1)$ with

2. 2
   $$f'(c)=\frac{f(1)-f(1/2)}{1-1/2}=\frac{1/2}{1/2}=1.$$
3. 3

   Since $f'$ is strictly increasing, $f'(1/2)<f'(c)=1<f'(1)$.

Answer structure / marking notes

Content note