6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I guess you mean $\int_1^2f'(x)dx$ instead of $\int_1^2f(x)dx$ $\endgroup$ – user378947 Nov 2 '16 at 22:20
  • $\begingroup$ @mathbeing yep. Thanks for pointing that out. $\endgroup$ – Jazz Nov 2 '16 at 22:21
  • $\begingroup$ 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. $\endgroup$ – Piotr Benedysiuk Nov 2 '16 at 22:25
5
$\begingroup$

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: $$ \frac{1}{2} \alpha^2 f'(\alpha) = f(2) - f(1) $$

$\endgroup$
2
$\begingroup$

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 $(1,2)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.