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Let $f:\mathbb R \to \mathbb R$ be a twice continuously differentiable function, with $f(0)=f(1)=f'(0) = 0$. Then

  1. $f^{"}$ is the zero function.

  2. $f^{"}(0)$ is zero.

  3. $f^{"}(x)=0$ for some $x \in $ (0,1).

  4. $f^{"}$ never vanishes.

I don't know where to start, tried rolles theorem but cant catch the point..please help me to start..

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  • $\begingroup$ I take it you're supposed to determine which of these things happens? More than one is possible. $\endgroup$ – Jakob Hansen Apr 15 '16 at 15:48
  • $\begingroup$ There is one that will happen in all cases, though. $\endgroup$ – Mauro Apr 15 '16 at 15:50
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    $\begingroup$ A fully satisfying answer should select the correct answer, AND show by way of counterexample or in any other manner why the other three answers are not correct too. $\endgroup$ – mathguy Apr 15 '16 at 15:56
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Rolle's theorem says that because $f(0) = f(1)$, there is a point $t$ STRICTLY between 0 and 1 where $f'(t) = 0$. Now use this and $f'(0) = 0$ and apply Rolle's theorem to $f'$ over $[0,t]$. This will lead you to the correct choice.

For practice, try to understand why the other three offered answers are not also correct (why the problem has only ONE correct choice).

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  • $\begingroup$ Got it..The answer is 3rd one...Thanks guys.. $\endgroup$ – Sam Christopher Apr 16 '16 at 12:49
  • $\begingroup$ 4th option is not possible.now take f(x)=x^3-x^2 the counter example for 1st and 2nd option. $\endgroup$ – Sam Christopher Apr 16 '16 at 12:54
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Hoping someone may find it useful: Let $$f(x)=\cos2\pi x-1, f(0)=f(1)=0$$ $$f'(x)=2\pi\sin2\pi x,f'(0)=0$$

It's easy to counter 1,2 and 4 with $f$ ($f''$ vanishes at $x=\frac14$). And for 4 alone, $f(x)=0$ works.

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