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

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

I really would like some hint. I noticed that the equation can be written

$$\int_1^2f(x)'\mathbb{d}x = f'(\alpha)\int_0^{\alpha} x\mathbb{d}x$$

EDIT: I confused the theorems. I guess I have to apply the Mean Value Theorem, but I don't know how.

• I guess you mean $\int_1^2f'(x)dx$ instead of $\int_1^2f(x)dx$ – user378947 Nov 2 '16 at 22:20
• @mathbeing yep. Thanks for pointing that out. – Jazz Nov 2 '16 at 22:21
• First thing coming to my mind is to work with some other function $g(x)$ defined using $f(x)$ that is also continuous, and use the Mean Value Theorem there. – Piotr Benedysiuk Nov 2 '16 at 22:25

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 $(1,2)$