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 $f'(x) = 0$
$(x - 2)^4 = 0 \Rightarrow x = 2$
$(x + 2) = 0 \Rightarrow x = -2$
$7x + 6 = 0 \Rightarrow x = -\frac{6}{7}$
Second Derivative:
Use derivative of $f'(x) = (x - 2)^4(x + 2)(7x + 6)$
At $x = -2$:
Since $(x + 2)^2$ in $f(x)$ has even power, and $f''(-2) < 0$ (can be verified), it's a local maximum.
At $x = -\frac{6}{7}$:
$f''(-\frac{6}{7}) > 0$ ⇒ local minimum
At $x = 2$:
$(x - 2)^5$ ⇒ odd power, and $f''(2) = 0$, sign of $f'(x)$ does not change ⇒ point of inflection
Answer: Point of local maxima is x = -2
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