Question

Let $f\left(x\right)=x^3+3x^2+3x+2$. Then at $x=-1$

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

Answer 1

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

$f\left(x\right)=x^3+3x^2+3x+2$
$f^{\prime }\left(x\right)=3x^2+6x+3=3(x^2+2x+1)=3(x+1{)}^2$
For minimum or maximum value of $f\left(x\right)$
$f^{\prime }\left(x\right)=0=3(x+1{)}^2\Rightarrow x=-1$
Now $f^{\prime \prime }\left(x\right)=6(x+1)\Rightarrow f^{\prime \prime }(-1)=0$ so we can't say anything about maxima or minima of $f\left(x\right)$
And for minimum or maximum value of $f^{\prime }\left(x\right)$
$f^{\prime \prime }\left(x\right)=0=6(x+1)\Rightarrow x=-1$
Now $f^{\prime \prime \prime }\left(x\right)=6>0$
Hence at $x=-1$ $f^{\prime }\left(x\right)$ achieves it's minimum value