Question

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a thrice differentiable odd function satisfying $f^{\prime}(x) \geq 0, f^{\prime\prime}(x)=f(x), f(0)=0, f^{\prime}(0)=3$. Then $9 f\left(\log _e 3\right)$ is equal to __________ .

Hint:

Solve $f''=f$ with the odd-function condition and $f'(0)=3$ to get $f(x)$, then evaluate it at $x=\log_e 3$.

Answer 1

Correct Answer: 36

Since $f''(x) = f(x)$, the general solution is $f(x) = Ae^{x} + Be^{-x}$.
$f$ is odd and $f(0)=0$, so $A + B = 0$ and $f(x) = A(e^{x} - e^{-x})$.
$f'(x) = A(e^{x} + e^{-x})$ and $f'(0) = 2A = 3$, so $A = \frac{3}{2}$.
Thus $f(x) = \frac{3}{2}(e^{x} - e^{-x})$, and $f'(x) = \frac{3}{2}(e^{x}+e^{-x}) > 0$ satisfies the condition $f'(x) \geq 0$.
At $x = \log_e 3$: $f(\log_e 3) = \frac{3}{2}\left(3 - \frac{1}{3}\right) = \frac{3}{2} \cdot \frac{8}{3} = 4$.
Therefore $9f(\log_e 3) = 36$.