If $f$ is differentiable on $[1,2]$, then $\exists \alpha\in(1,2): f(2)-f(1) = \frac{\alpha^2}{2}f'(\alpha)$

Let $ g(x) = f(1/x) $ for $ x \in [1/2, 1] $. Applying the mean value theorem, there is a $ c \in (1/2, 1)$ with $$ g'(c) = \frac{g(1) - g(1/2)}{1/2} = 2 \; ( g(1) - g(1/2) ) $$ Rewrite this in terms of $ f $, using $ g'(c) = -\frac{1}{c^2} f'\left(\frac{1}{c} \right) $, and $ g(1) = f(1) $, and $ g(1/2) = f(2) $, to find: $$ \frac{1}{c^2} f'\left(\frac{1}{c} \right) = 2 (f(2) - f(1) ) $$ Set $\alpha = 1/c $ so that $ \alpha \in (1,2) $ and divide by $ 2 $ to find: $$ \frac{1}{2} \alpha^2 f'(\alpha) = f(2) - f(1) $$

Hint: Consider the function $g(x)= f'(x)- \frac{2(f(2)-f(1))}{x^2}$ in the interval $[1,2]$ Now if you integrate you'll find that $G(x)=\int_1^x g(t)dt= f(x)+\frac{2(f(2)-f(1))}{x}+f(1)-2f(2)$ Now it is easy to see that $ G(1)=G(2)=0$ so according to Rolle's theorem there is a root of $G'(=g)$ in $(1,2)$