Question

lf $f\left(x\right)=x^3+a{x}^2+bx+c$ has a maximum at $x=-1$ and minimum at $x=3$, then

• A. $f(-1)<f\left(3\right)$
• B. $a=-3$
• C. $b=-9$
• D. $c\in (-11,27)$

Answer 1

Answer: (B), (C)

The correct options are
B $a=-3$
C $b=-9$

$f\left(x\right)=x^3+a{x}^2+bx+c$
$\Rightarrow f^{\prime }\left(x\right)=3x^2+2ax+b$
Also $f^{\prime \prime }\left(x\right)=6x+2a$
Since we have a maximum at $x=-1$, $f^{\prime }(-1)=0$ and $f^{\prime \prime }(-1)<0$
$\Rightarrow 3-2a+b=0$ and $-6+2a<0$
Since there is a minimum at $x=3,f^{\prime }\left(3\right)=0$ and $f^{\prime \prime }\left(3\right)>0$
$\Rightarrow 27+6a+b=0$ and $18+2a>0$

by solving the above equations we get
$\Rightarrow a=-3,b=-9,c\in R$
Hence, options B,C become correct.