Question

Let $f\left(x\right)=x^3+6x^2+12x+15\forall xϵR$, at $x=-2$

• A. $f\left(x\right)$ has a maximum
• B. $f\left(x\right)$ has minimum
• C. $f^{\prime }\left(x\right)$ has a maximum
• D. $f^{\prime }\left(x\right)$ has minimum

Answer 1

The correct option is D $f^{\prime }\left(x\right)$ has minimum

$f\left(x\right)=x^3+6x^2+12x+15$
$\Rightarrow f^{\prime }\left(x\right)=3x^2+12x+12=3(x+2{)}^2$
$\Rightarrow f^{\prime \prime }\left(x\right)=6(x+2)$
$\Rightarrow f^{\prime \prime \prime }\left(x\right)=6$
clearly at $x=-2,f^{\prime }\left(x\right)=0=f^{\prime \prime }\left(x\right)$ and $f^{\prime \prime \prime }\left(x\right)>0$
Hence only conclusion drawn from here is that $f^{\prime }\left(x\right)$ has minima at $x=-2$