Question

Let $f:R\to R$ be a twice differentiable function such that $f(a)=0,f(b)=2,f(c)=-1,f(d)$ $=2,f(e)=0$, where $a<b<c<d<c$. Determine the minimum number of zeroes of $g(x)={\left(f^{\prime }(x)\right)}^2+f(x)\cdot f^{\prime \prime }(x)$ in the interval $[a,e]$.

Answer 1

Answer: 6

$g(x)={\left(f^{\prime }(x)\right)}^2+f(x)\cdot f^{\prime \prime }(x)$
$\therefore g(x)=\frac{d}{dx}\left(f(x)\cdot f^{\prime }(x)\right)$
$f(a)=f(c)=0$
Since $f$ is a continuous function and $f(b)\&f(x)$ have opposite sign $\therefore$ Using IVT $\exists$ atleast one $\alpha$ root in $(b,c)$ similarly $f(d)$ and $f(x)$ have opposite sign. $\therefore \exists$ atleast one $\beta$ root in $(c,d)$.
$\therefore f(x)=0$ has atleast 4 real roots in $[a,e]$
Now $f(a)=f(\alpha )=0$ and $f$ is a continuous and derivable function in $[a,\alpha ]$
$\therefore$ using Rolle's Theorem $\exists$ atleast some $C_1$ in $[a,\alpha ]$ where $f^{\prime }\left(c_1\right)=0$
again using Rolle's in $(\alpha ,\beta )\exists$ some $c_2$ in $(\alpha ,\beta )$ where $f^{\prime }\left(c_2\right)=0\Rightarrow f^{\prime }(x)$ atleast has 3 real roots.
Using Rolle's in $(\beta ,e)$
$\exists$ some $c_3$ in $(\beta ,e)$ where $f^{\prime }\left(c_3\right)=0$
$\therefore$ The function $f^{\prime }x)\cdot f^{\prime }(x)$ has atleast 7 real roots.
$\therefore g(x)=\frac{d}{dx}\left(f(x)\cdot f^{\prime }(x)\right)$ has atleast 6 real roots.