Question

Let $f\left(x\right)$ be a cubic polynomial with $f\left(1\right)=-10,$ $f(-1)=6,$ and has a local minima at $x=1$, and $f^{\prime }\left(x\right)$ has a local minima at $x=-1.$ Then $f\left(3\right)$ is equal to

Answer 1

$f\left(1\right)=-10,f(-1)=6$
$f^{\prime }\left(1\right)=0$ and $f^{\prime \prime }(-1)=0$ as given
$f\left(x\right)$ has minima at $x=1$
and $f^{\prime }\left(x\right)$ has minima at $x=-1$
So, $f^{\prime \prime }\left(x\right)=a(x+1)$
Integrating both sides
$f^{\prime }\left(x\right)=\frac{a}{2}(x+1{)}^2+c$
$f^{\prime }\left(1\right)=0=2a+c$
$\Rightarrow c=-2a$
$f^{\prime }\left(x\right)=\frac{a}{2}(x+1{)}^2-2a$
Intergrating both side
$f\left(x\right)=\frac{a}{6}(x+1{)}^3-2xa+c^{\prime }$
$f\left(1\right)=-10=\frac{8a}{6}-2a+c^{\prime }$
$f(-1)=6=2a+c^{\prime }$
$2a+c^{\prime }=6$
$4a-6c^{\prime }=60$
$\Rightarrow a=6,c^{\prime }=-6$
$f\left(x\right)=(x+1{)}^3-12x-6$
$f\left(3\right)=(4{)}^3-36-6$
$f\left(3\right)=22$