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$f(x)$ has second derivative on $[0,1]$, and $f(0)=f'(0), f(1)=f'(1)$. Prove that: there exists $\xi\in (0,1)$, s.t. $f(\xi)=f''(\xi)$.

Let $g(x)=\left(f(x)\right)^2-\left(f'(x)\right)^2$, then $g(0)=g(1)=0$. By Rolle's theorem there exists $\xi\in (0,1)$, s.t. $g'(\xi)=2f'(\xi)(f(\xi)-f''(\xi))=0$. If $f'(\xi) \neq 0$ it will work. However, what if $f'(\xi)=0$?

I found a problem almost the same, just adding a condition $f(0)=f'(0)=f(1)=f'(1)=0$. With this condition, we consider a function $h(x)=\dfrac{f(x)+f'(x)}{\mathrm{e}^x}$. Applying Rolle's theorem there exists $\xi\in (0,1)$, s.t. $h'(\xi)=\dfrac{f''(\xi)-f(\xi)}{\mathrm{e}^\xi}=0$, thus satisfying the requirement.

So is the original problem correct without the condition?

Saunders
  • 656

1 Answers1

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Take $h(x) =(f(x) - f' (x)) e^x $ then $$h' (x) =(f'(x) -f'' (x)e^x +e^x (f(x)-f'(x)) =(f(x)-f'' (x))e^x $$ So you can use a Rolle theorem since $h(1) =h(0)=0.$