Question

Let $f\left(x\right)$ be a polynomial of degree $3$ such that $f(-1)=10,f\left(1\right)=6,f\left(x\right)$ has a critical point at $x=-1$ and $f^{\prime }\left(x\right)$ has a critical point at $x=1$. Then $f\left(x\right)$ has a local minima at $x$ equals to

Answer 1

Let the polynomial be
$f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$\Rightarrow f^{\prime }\left(x\right)=3a{x}^2+2bx+c$
$\Rightarrow f^{\prime \prime }\left(x\right)=6ax+2b$
$f^{\prime \prime }\left(1\right)=0\Rightarrow 6a+2b=0\Rightarrow b=-3a$
$f^{\prime }(-1)=0\Rightarrow 3a-2b+c=0$
$\Rightarrow c=-9a$
$f(-1)=10\Rightarrow -a+b-c+d=10$
$\Rightarrow -a-3a+9a+d=10$
$d=-5a+10$
$f\left(1\right)=-6\Rightarrow a+b+c+d=-6$
$\Rightarrow a-3a-9a-5a+10=-6$
$\Rightarrow a=1$
$\therefore f^{\prime }\left(x\right)=3x^2-6x-9=3(x^2-2x-3)$
For $f^{\prime }\left(x\right)=0\Rightarrow x^2-2x-3=0\Rightarrow x=3,-1$
Minima exists at $x=3$