Question

Local maximum and local minimum values of the function $\left(x-1\right){\left(x+2\right)}^2$ are

• A. $-4,0$
• B. $0,-4$
• C. $4,0$
• D. None of these

Answer 1

The correct option is B

$0,-4$

Explanation for Correct Answer:

Finding the local maximum and local minimum:

Let

$$\begin{gathered} f\left(x\right)=\left(x-1\right){\left(x+2\right)}^2 \\ f\text{'}\left(x\right)=2(x-1)(x+2)+{(x+2)}^2 \\ =2x^2-2x+4x-4+x^2+4x+4 \\ =3x^2+2x+4x \\ =3x^2+6x \end{gathered}$$

equating $f\text{'}\left(x\right)=0$,

$$\begin{gathered} 3x^2+6x=0 \\ 3x(x+2)=0 \\ \Rightarrow x=0;x=-2 \end{gathered}$$

These are the critical points.

Substitute these point in $f\left(x\right)$.

$f\left(0\right)=\left(0-1\right){\left(0+2\right)}^2=-4<0$ is the minimum value of $f\left(x\right)$.

$f\left(1\right)=\left(1-1\right){\left(1+2\right)}^2=0$ is the maximum value of $f\left(x\right)$

Local maximum and local minimum values of $f\left(x\right)$ is $0\text{and}-4$

Hence, the correct answer is option (B).