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