Find the local maxima and local minima, if any, of the function $ g(x) = x^3 - 3x $.

Q: Find the local maxima and local minima, if any, of the function $ g(x) = x^3 - 3x $.

Explanation: \[ g(x) = x^3 - 3x \Rightarrow g'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1) \] Setting $ g'(x) = 0 \Rightarrow x = \pm 1 $ \[ g''(x) = 6x \Rightarrow g''(1) = 6 > 0 \Rightarrow \text{Local minimum at } x = 1 \] \[ g(1) = 1 - 3 = -2 \] \[ g''(-1) = -6 < 0 \Rightarrow \text{Local maximum at } x = -1 \] \[ g(-1) = -1 + 3 = 2 \]

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Explanation: $$
g(x) = x^3 - 3x \Rightarrow g'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)
$$
Setting $ g'(x) = 0 \Rightarrow x = \pm 1 $

$$
g''(x) = 6x
\Rightarrow g''(1) = 6 > 0 \Rightarrow \text{Local minimum at } x = 1
$$
$$
g(1) = 1 - 3 = -2
$$
$$
g''(-1) = -6 < 0 \Rightarrow \text{Local maximum at } x = -1
$$
$$
g(-1) = -1 + 3 = 2
$$

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