For every twice differentiable function f : R | vedclass

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(D) Given $f : R \rightarrow [-2, 2]$ and $(f(0))^2 + (f'(0))^2 = 85$. Since the codomain is $[-2, 2]$,the function cannot be constant. Thus,$f(x)$ mu

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(D) Given $f : R ightarrow [-2, 2]$ and $(f(0))^2 + (f'(0))^2 = 85$. Since the codomain is $[-2, 2]$,the function cannot be constant. Thus,$f(x)$ must be increasing or decreasing in some small interval,implying there exist $r, s \in R$ with $r $(B)$ Applying the Mean Value Theorem $(LMVT)$ on the interval $[-4, 0]$,there exists $x_0 \in (-4, 0)$ such that $f'(x_0) = \frac{f(0) - f(-4)}{4}$. Since $|f(x)| \leq 2$,we have $|f'(x_0)| = \left| \frac{f(0) - f(-4)}{4} ight| \leq \frac{2 + 2}{4} = 1$. Thus,$|f'(x_0)| \leq 1$.$(C)$ Let $f(x) = \sin(\sqrt{85}x)$. Then $f(0) = 0$ and $f'(x) = \sqrt{85}\cos(\sqrt{85}x)$,so $f'(0) = \sqrt{85}$. However,$\lim_{x ightarrow \infty} \sin(\sqrt{85}x)$ does not exist. Thus,$(C)$ is incorrect.$(D)$ Define $g(x) = (f(x))^2 + (f'(x))^2$. Then $g'(x) = 2f(x)f'(x) + 2f'(x)f''(x) = 2f'(x)(f(x) + f''(x))$. Since $g(0) = 85$ and $|f(x)| \leq 2$ and $|f'(x)| \leq 1$ (by $LMVT$ logic),$g(x)$ must have a maximum in $(-4, 4)$. At the maximum $\alpha$,$g'(\alpha) = 0$. Since $f'(x)$ cannot be zero everywhere,there exists $\alpha$ such that $f'(\alpha) 
eq 0$ and $f(\alpha) + f''(\alpha) = 0$.

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