For a twice differentiable function $f(x), g(x)$ is defined as $g(x)=\left(f^{\prime}(x)^{2}+f^{\pri

Question

For a twice differentiable function $f(x), g(x)$ is defined as $g(x)=\left(f^{\prime}(x)^{2}+f^{\prime \prime}(x)\right) f(x)$ on $[a, e]$. If for $a<b<c<d<e$, $f(a)=0, f(b)=2, f(c)=-1, f(d)=2, f(e)=0$ then find the minimum number of zeros of $g(x)$.

Professor PC Thomas Classes · Answer

$\frac{d}{d x}\left(f(x) f^{\prime}(x)ight)=f^{\prime \prime}(x) f(x)+\left(f^{\prime}(x)ight)^{2}$ $=g(x)$\ Let $\phi(x)=f(x) f^{\prime}(x)$\ $f(a)=0, f(b)=2$ and $f(c)=-1$\ Since $f$ is continuous $\exists k_{1}$ such that\ $f\left(k_{1}ight)=0$\ $f(d)=2 \exists k_{2}$ such that $f\left(k_{2}ight)=0$\ $f(e)=0$\ $	herefore \quad f(x)=0$\ $	herefore \quad f$ has a minimum of four roots. Between any two roots of a polynomial equation $\exists$ a root of $f^{\prime}(x)=0$\ $	herefore \quad$ There are at least seven roots for $\phi(x)=0$\ $	herefore \quad g(x)$ has at least six roots.\