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

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

To find critical points, find the first derivative using the product rule:

Let $u = (x - 2)^5$ and $v = (x + 2)^2$

$u' = 5(x - 2)^4$

$v' = 2(x + 2)$

$f'(x) = 5(x - 2)^4(x + 2)^2 + (x - 2)^5 \cdot 2(x + 2)$

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Factor out the common terms $(x - 2)^4$ and $(x + 2)$:

$f'(x) = (x - 2)^4(x + 2)[5(x + 2) + 2(x - 2)]$

$f'(x) = (x - 2)^4(x + 2)[5x + 10 + 2x - 4]$

$f'(x) = (x - 2)^4(x + 2)(7x + 6)$

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Set $f'(x) = 0$:

$(x - 2)^4(x + 2)(7x + 6) = 0$

Critical points are:

$x = 2$

$x = -2$

$x = -\dfrac{6}{7}$

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Apply the first derivative test. Note that $(x - 2)^4$ is always positive.

For $x < -2$, test $x = -3$:

$(x + 2) < 0$ and $(7x + 6) < 0$, so $f'(x) > 0$ (increasing)

For $-2 < x < -\dfrac{6}{7}$, test $x = -1$:

$(x + 2) > 0$ and $(7x + 6) < 0$, so $f'(x) < 0$ (decreasing)

At $x = -2$, the function changes from increasing to decreasing.

Therefore, $x = -2$ is a local maximum.