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

Question

Let $f:(0,\infty)	o R$ be a twice differentiable function. If for some $a
e 0,\int _{0}^{1}f(\lambda x)d\lambda =af(x),f(1)=1$ and $f(16)=\frac{1}{8}$, then $16-{f}^{'}(\frac{1}{16})$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 112

Accepted Answer

$\begin{gathered} {\int }_{1}^{0}f(\lambda x)d\lambda =af(x) \ \lambda x=t \ d\lambda =\frac{1}{x}dt \ \frac{1}{x}{\int }_{x}^{0}f(t)dt=af(x) \ {\int }_{x}^{0}f(t)dt=axf(x) \ f(x)=a(x{f}^{'}(x)+f(x)) \ (1-a)f(x)=a\cdot x{f}^{'}(x) \ \frac{{f}^{'}(x)}{f(x)}=\frac{(1-a)}{a}\frac{1}{x} \ llnf(x)=\frac{1-a}{a}lnx+c \ x=1,f(1)=1\Rightarrow c=0 \ x=16,f(16)=\frac{1}{8} \end{gathered}$

$\begin{gathered} \frac{1}{8}=(16{)}^{\frac{1-a}{a}}\Rightarrow -3=\frac{4-4a}{a}\Rightarrow a=4 \ f(x)={x}^{-\frac{3}{4}} \ {f}^{'}(x)=-\frac{3}{4}{x}^{-\frac{7}{4}} \ 	herefore \, 16-{f}^{'}(\frac{1}{16}) \ =16-(-\frac{3}{4}{({2}^{-4})}^{-7/4}) \ =16+96=112 \end{gathered}$

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