The point of local maxima of the function $f(x) = (x-2)^5(x+2)^2$ is

-2

The correct answer is Option (4) → -2

Given: $f(x) = (x - 2)^5(x + 2)^2$

First Derivative:
$f'(x) = \frac{d}{dx}[(x - 2)^5(x + 2)^2]$
Using product rule:
$f'(x) = (x - 2)^4(x + 2)[5(x + 2) + 2(x - 2)]$
$= (x - 2)^4(x + 2)(7x + 6)$

Critical Points:
Set $f'(x) = 0$
$(x - 2)^4 = 0 \Rightarrow x = 2$
$(x + 2) = 0 \Rightarrow x = -2$
$7x + 6 = 0 \Rightarrow x = -\frac{6}{7}$

Nature of critical points (using sign change of f′(x))
 $f'(x) = (x - 2)^4(x + 2)(7x + 6)$

At $x = -2$: $f'(x)$ changes from positive to negative, hence it is a local maximum.

At $x = -\frac{6}{7}$:$f'(x)$ changes from negative to positive, hence it is a local minimum.

At $x = 2$: $(x - 2)^4$ is an even power , so $f'(x)$ does not change sign ⇒ point of inflection.

Answer: Point of local maxima is x = -2