Question

$f\left(x\right)=x^3+a{x}^2+bx+c$ has a max. at $x=-1$ and min. at $x=3$. Determine the constants $a,b,c$.

• A. $\therefore a=-3,b=-3,c=0.$
• B. $\therefore a=-3,b=-9,c=3.$
• C. $\therefore a=-3,b=9,c=9.$
• D. $\therefore a=3,b=9,c=-3.$

Answer 1

The correct option is B $\therefore a=-3,b=-9,c=3.$

Given $y=x^3+a{x}^2+bx+c$
$\frac{dy}{dx}=3x^2+2ax+b=0$

At $x=-1$
$\therefore 3-2a+b=0$

$\therefore$ $-2a+b=-3$ ....(1)

At $x=3$
$27+6a+b=0$

$\therefore$ $6a+b=-27$ ....(2)

Solving (1) and (2), we get
$a=-3,b=-9$
There is no condition for $c$, therefore $c$ can assume any real value.
$\therefore a=-3,b=-9,cϵR.$