Let f:(0,)R be a twice differentiable function.

Question

Let $f:(0,\infty)	o\mathbb{R}$ be a twice differentiable function. If $\displaystyle\int_0^1 f(\lambda x)\,d\lambda = af(x)$ for some $a
e0$, and $f(1)=1$, then $16-f'\left(\dfrac1{16}ight)$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 112

Accepted Answer

Substitute $\lambda x = t$: $\dfrac1x\displaystyle\int_0^x f(t)\,dt = af(x) \implies \int_0^x f(t)\,dt = axf(x)$.

Differentiating: $f(x) = a[xf'(x)+f(x)] \implies (1-a)f(x) = ax f'(x) \implies \dfrac{f'(x)}{f(x)} = \dfrac{1-a}{a}\cdot\dfrac1x$

Integrating: $\ln f(x) = \dfrac{1-a}{a}\ln x + c$. Using $f(1)=1$: $c=0$.

Given $f(16) = \dfrac18$ (used as auxiliary condition): $\dfrac18 = 16^{(1-a)/a} \implies -3 = \dfrac{4-4a}{a} \implies a=4$.

So $f(x) = x^{-3/4}$, $f'(x) = -\dfrac34 x^{-7/4}$.

$$f'\left(\frac1{16}\right) = -\frac34(16)^{7/4} = -\frac34(2^7) = -96$$

$$16 - f'\left(\frac1{16}\right) = 16+96 = 112$$

Solutions · 1