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let $f(x)$ is continuous on $[0,1/2]$, and derivative on $(0,1/2)$,such $$f'(1/2)=0$$ show that

there exsit $c\in(0,1/2)$, such $$f'(c)=2c(f(c)-f(0))$$

My try: let $$F(x)=e^{-x^2}[f(x)-f(0)]\Longrightarrow F'(x)=e^{-x^2}[f'(x)-2x(f(x)-f(0))]$$ then I can't

mathlove
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2 Answers2

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I dont think the above answer is not accurate. If one wants to use that method, a little more work should be done with the sign of $g(x)$ by stating this:
If $h$ is a differential function on [a,b] and $h'(a)\le h'(a)( a<b)$ hence: $\forall \lambda \in [h'(a);h'(b)]: \exists c \in [a,b]: h'(c)=\lambda$
Another approach:
Denote $F(x)$ as defined be by nanchangjian.
Obviously, $F(x)$ has its maximum. WLOG: $F(c)=max(F(x))(c \in [0;\frac{1}{2}])$.
If $ 0<c<\frac{1}{2}$, the conclusion becomes obvious.
If $c=\frac{1}{2}$, thus $ F'(\frac{1}{2}) \ge 0 \Leftrightarrow f(0) \ge f(\frac{1}{2})$
$ \Rightarrow F(0) \ge F(1/2) \Rightarrow QED$
If $c=0$ , we just have to consider the minimum $F(d)$ of $F(x)$.

Toan Nguyen Dinh
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Suppose,contrary to our claim，Use the Darboux theorem： http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)

Without loss of , we Assmue that $$g(x)=f'(x)-2x[f(x)-f(0)]>0,0<x<1/2$$ then $$F(x)=e^{-x^2}[f(x)-f(0)]\Longrightarrow F'(x)=e^{-x^2}[f'(x)-2x(f(x)-f(0))]>0$$ so $$F(x)>F(0)=0$$ so $$f'(x)-2x(f(x)-f(0))>0,0<x\le\dfrac{1}{2}$$ let $x=1/2$ then $$g(1/2)=f'(1/2)-(f(1/2)-f(0))=-(f(1/2)-f(0))<0$$ a contradiction.

math110
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  • "Without loss of"...gnerality? How? Why couldn't $;g(x);$ take positive and negative values? After all we're not given $;f';$ is continuous... – DonAntonio Jan 06 '14 at 12:07
  • if $g(x)<0$ simaler methods can have contradiction – math110 Jan 06 '14 at 12:16
  • Come again, @math110 : "simaler" ...? Did you mean "similar"? But this doesn't address my question in my past comment: what if $;g;$ gets both positive and negative values? – DonAntonio Jan 06 '14 at 12:19
  • yse,this "similar", the Darboux theorem： http://en.wikipedia.org/wiki/Darboux's_theorem_(analysis) – math110 Jan 06 '14 at 12:25
  • Nice...then you should probably mention this in your answer, @math110 – DonAntonio Jan 06 '14 at 12:26
  • There is some minor issue here. You start with $g(x) > 0$ for $0 < x < 1/2$ and then at the end you write $g(1/2) > 0$ (second last equation) and then finally $g(1/2) < 0$ to get contradiction. The conclusion $g(1/2) > 0$ does not follow from anywhere. We can argue like this: since $g(x)$ satisfies IVT, we assume $g(x) > 0$ for all $x \in (0, 1/2)$. From this we show $F(1/2) > 0$ and then $g(1/2) < 0$. By IVT we must have $g(c) = 0$ somewhere in $(0, 1/2)$ and this contradicts that $g(x) > 0$ in $(0, 1/2)$. – Paramanand Singh Mar 02 '14 at 07:21