Question

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

Answer 1

Answer: 1

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