Question

If $f\left(x\right)$ is a twice differentiable function such that $f\left(a\right)=0,f\left(b\right)=2,f\left(c\right)=-1,f\left(d\right)=2,f\left(e\right)=0$ when $a<b<c<d<e$ then the minimum number of zeros of $g\left(x\right)={\left(f\right(x\left)\right)}^2+f^{\prime }\left(x\right)f\left(x\right)$ in the interval $[a,e]$

Answer 1

$g\left(x\right)=(f^{\prime }({x)}^2+f"(x\left)f\left(x\right)\right)$

$=(f^{\prime }(x{)}^2+f(x\left)f^{\prime \prime }\left(x\right)\right)$

$Bydifferentiatingd/dx$

$g\left(x\right)=\frac{d}{dx}\left(f\right(x\left)f^{\prime \prime }\left(x\right)\right)$

$h\left(x\right)=\left(f\right(x\left)f^{\prime \prime }\left(x\right)\right)$

$f\left(x\right)hasminimumoffourzerosand{f}^{\prime }\left(x\right)hasminimumoffourzeros$

$h\left(x\right)=f\left(x\right)f^{\prime }\left(x\right)$

$Weknowthath\left(x\right)=0$

$f\left(x\right)f^{\prime }\left(x\right)=0$

$Sof\left(x\right)=4zerosand{f}^{\prime }\left(x\right)=3zeros,Sinceh\left(x\right)=7zerosand{h}^{\prime }\left(x\right)=6zeros$

$therefore$

$g\left(x\right)=h^{\prime }\left(x\right)$