The function $f(x) = x + \frac{a^2}{x}$, $a > 0$, $x \neq 0$ has a local maxima at

Given: $f(x) = x + \frac{a^2}{x}$, where $a > 0$ and $x \neq 0$

To find local maxima, find the critical points where $f'(x) = 0$ and test their nature.

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Finding the first derivative:

$f'(x) = 1 - \frac{a^2}{x^2}$

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Setting $f'(x) = 0$ to find critical points:

$1 - \frac{a^2}{x^2} = 0$

$1 = \frac{a^2}{x^2}$

$x^2 = a^2$

$x = \pm a$

The critical points are $x = a$ and $x = -a$.

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Finding the second derivative:

$f''(x) = \frac{2a^2}{x^3}$

For $f''(x) < 0$, the critical point is a local maxima.

For $f''(x) > 0$, the critical point is a local minima.

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At $x = a$:

$f''(a) = \frac{2a^2}{a^3}$

$f''(a) = \frac{2}{a}$

Since $a > 0$, $f''(a) > 0$

This is a local minima.

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At $x = -a$:

$f''(-a) = \frac{2a^2}{(-a)^3}$

$f''(-a) = \frac{2a^2}{-a^3}$

$f''(-a) = -\frac{2}{a}$

Since $a > 0$, $f''(-a) < 0$

This is a local maxima.

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Therefore, the function has a local maxima at $x = -a$.