Question:
If $f(x)=x^{3}+a x^{2}+b x+c$ has a maximum at $x=-1$ and minimum at $x=3$. Determine $a, b$ and $c$.
Solution:
We have,
$f(x)=x^{3}+a x^{2}+b x+c$
$\Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b$
As, $f(x)$ is maximum at $x=-1$ and minimum at $x=3$.
So, $f(-1)=0$ and $f(3)=0$
$\Rightarrow 3(-1)^{2}+2 a(-1)+b=0$ and $3(3)^{2}+2 a(3)+b=0$
$\Rightarrow 3-2 a+b=0 \quad \ldots$ (i)
and $27+6 a+b=0 \quad \ldots$ (ii)
(ii) - (i), we get
$27-3+6 a+2 a=0$
$\Rightarrow 8 a=-24$
$\Rightarrow a=-3$
Substituting $a=-3$ in (i), we get
$3-2(-3)+b=0$
$\Rightarrow 3+6+b=0$
$\Rightarrow b=-9$
And, $c \in \mathbf{R}$