$f : \mathbb R \to [-2,2]$ be a twice differentiable with $f(0)^2+ f'(0)^2=85$

Asked Jun 15 '18 at 17:05

Active Jun 15 '18 at 19:32

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Let $f : \mathbb R \to [-2,2]$ be a twice differentiable with $f(0)^2+ f'(0)^2=85$. Is it true that $\exists a \in (-4,4)$ such that $f'(a) \ne 0$ and $f(a)+f''(a)=0$ ?

I think I have to use IVT / MVT to some nicely constructed new function, but I am unable to see what. Please help.

asked Jun 15 '18 at 17:05

Do you mean $(-2,2)$? – Clayton Jun 15 '18 at 17:09
@Clayton: Well I am required to get $a$ in $(-4,4)$ , anyway if you can get $a$ in $(-2,2)$ then definitely that's in $(-4,4)$ ... – Jun 15 '18 at 17:27
@Clayton Why $(-2, 2)$? The function $f$ is defined on all of $\Bbb R$. – Arthur Jun 15 '18 at 18:06
@Arthur: Symmetry of the question (this is all I had in mind; I wasn't under the impression that $(-4,4)$ wasn't in the domain). – Clayton Jun 15 '18 at 19:31
Does this answer your question? For every twice differentiable function $f : \bf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statements are TRUE? – Calvin Khor Jan 27 '22 at 07:30

$f : \mathbb R \to [-2,2]$ be a twice differentiable with $f(0)^2+ f'(0)^2=85$

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