if f is differentiable on [a,b] and f' is decreases strictly on [a,b] prove that
f'(b)

Question

if f is differentiable on [a,b] and f' is decreases strictly on [a,b] prove that 
               f'(b)<(f(b)-f(a))/(b-a)<f'(a)

anonymous · Answer

I haven't seen the mean value theorem of calculus this way yet to be honest. However, let me write down what I know about it, maybe it will help you with your problem nevertheless. 

$$ f'(x_0)= \frac{f(b)-f(a)}{b-a} \ \text{if}  \ f'(x_0) < 0 \ \text{and}  \ b-a<0 \ \text{then} 
\ f'(x_0)(b-a) < 0 \
\therefore \ f(b)-f(a)<0 \
\therefore \ f(b)<f(a)  $$
Maybe this helps you with your problem.

anonymous · Answer

note that the second line is the definition of a strictly decreasing function.