Question

Let $f$ be a twice differentiable function such that $f^{\prime \prime }(x)=-f(x)$ and $f^{\prime }(x)=g(x)$, $h(x)=(f(x){)}^2+(g(x){)}^2$. If $h(17)=23$, then find $h(51)$.

Answer 1

Answer: 23

$h(x)=(f(x){)}^2+(g(x){)}^2$
Differentiate with respect to $x$, we get
$h^{\prime }(x)=2f(x)f^{\prime }(x)+2g(x)g^{\prime }(x)$
$=2f(x)g(x)+2g(x)g^{\prime }(x)$
$=2f(x)g(x)+2g(x)\left(f^{\prime \prime }(x)\right)$
$=2f(x)g(x)-2f(x)g(x)$
$=0$
$\Rightarrow h$ is a constant function
$h(17)=23$
$\Rightarrow h(x)=23$, for all $x$
$\Rightarrow h(51)=23$