Question:
Let $f(x)$ be a cubic polynomial with $f(1)=-10$, $f(-1)=6$, and has a local minima at $x=1$, and $f^{\prime}(x)$ has a local minima at $x=-1$. Then $f(3)$ is equal to
Solution:
$F^{\prime}(x)=a(x-1)(x+3)$
$F^{\prime \prime}(x)=6 a(x+1)$
$F^{\prime}(x)=3 a(x+1)^{2}+b$
$F^{\prime}(1)=0 \Rightarrow b=-12 a$
$F(x)=a(x+1)^{3}-12 a x+c$
$=(x+1)^{3}-12 x-6$
$\mathrm{F}(3)=64-36-6=22$