Analysis (MVT)

• Feb 26th 2010, 01:33 PM
Aidan1930
Analysis (MVT)
This question is driving me mad.

Let f be continuous on [−1, 1] and twice differentiable on (−1, 1). Let φ(x) = (f (x) − f (0))/x for x ̸= 0 and φ(0) = f ′ (0). Show that φ is continuous on [−1, 1] and differentiable on (−1, 1). *Using a second order mean value theorem for f, show that φ′(x) = f′′(θx)/2 for some 0 < θ < 1 *. Hence prove that there exists c ∈ (−1, 1) with f ′′ (c) = f (−1) + f (1) − 2f (0).

It's the starred bit I can't do. I'm getting so frustrated. Any hints would be appreciated. Thanks
• Feb 26th 2010, 01:53 PM
Drexel28
Quote:

Originally Posted by Aidan1930
This question is driving me mad.

Let f be continuous on [−1, 1] and twice differentiable on (−1, 1). Let φ(x) = (f (x) − f (0))/x for x ̸= 0 and φ(0) = f ′ (0). Show that φ is continuous on [−1, 1] and differentiable on (−1, 1). *Using a second order mean value theorem for f, show that φ′(x) = f′′(θx)/2 for some 0 < θ < 1 *. Hence prove that there exists c ∈ (−1, 1) with f ′′ (c) = f (−1) + f (1) − 2f (0).

It's the starred bit I can't do. I'm getting so frustrated. Any hints would be appreciated. Thanks

Clearly $\displaystyle \varphi'(x_0)=\lim_{x\to x_0}\frac{\varphi(x)-\varphi(x_0)}{x-x_0}=?$

It is clear for $\displaystyle x=0$ since $\displaystyle \varphi'(0)=\lim_{x\to0}\frac{\varphi(x)-f'(0)}{x}=\lim_{x\to 0}\frac{f(x)-f(0)-x\cdot f'(0)}{x^2}$ and since $\displaystyle f$ is differentiable we are allowed to use L'hopitals rule to get $\displaystyle \lim_{x\to0}\frac{f'(x)-f'(0)}{2x}=\frac{f''(0)}{2}$.

Try the rest yourself.