Question

The differentiable functions $f, g$ and $h$ are such that $f'(x)=g(x), g'(x)=h(x), h'(x)=f(x)$, $f(0)=1, g(0)=0=h(0)$, find $[f(x)]^{3}+[g(x)]^{3}+[h(x)]^{3}-3 f(x) g(x) h(x)$ at $x=7$.

Answer 1

Let $y(x)=[f(x)]^{3}+[g(x)]^{3}+[h(x)]^{3} -3 f(x) g(x) h(x)$
$\Rightarrow y'(x)=3[f(x)]^{2} f'(x) +3[g(x)]^{2} g'(x) +3[h(x)]^{2} h'(x)-3\left[f(x) g(x) h'(x)\right.\left. +f(x) g'(x) h(x)+f'(x) g(x) h(x)\right]$
$=3[f(x)]^{2} g(x)+3[g(x)]^{2} h(x)+3[h(x)]^{2} f(x) -3[f(x) g(x) f(x)+f(x) h(x) h(x)+g(x) g(x) h(x)]$
$=0$
$\Rightarrow y(x)$ is a constant for all $x \in R$
$\Rightarrow y(7)=y(0)$
$y(0)=[f(0)]^{3}+[g(0)]^{3}+[h(0)]^{3}-3 f(0) g(0) h(0)$
$=1^{3}+0+0-3(0)$
$=1$
$\Rightarrow y(7) =1$