Find the absolute maximum and minimum values of the function $ f(x) = x^3 $ in the interval $[ -2, 2 ]$.

Q: Find the absolute maximum and minimum values of the function $ f(x) = x^3 $ in the interval $[ -2, 2 ]$.

Explanation: We have $ f(x) = x^3 $ \[ f'(x) = 3x^2 \] Setting $ f'(x) = 0 \Rightarrow x = 0 $ Now evaluate $ f $ at endpoints and critical points: \[ f(-2) = (-2)^3 = -8 \\ f(0) = 0 \\ f(2) = 2^3 = 8 \] $\textbf{Absolute maximum}$ is $ f(2) = 8 $ at $ x = 2 $ and $\textbf{Absolute minimum}$ is $ f(-2) = -8 $ at $ x = -2 $

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Explanation: We have $ f(x) = x^3 $  
$$
f'(x) = 3x^2
$$
Setting $ f'(x) = 0 \Rightarrow x = 0 $

Now evaluate $ f $ at endpoints and critical points:
$$
f(-2) = (-2)^3 = -8 \
f(0) = 0 \
f(2) = 2^3 = 8
$$

$\textbf{Absolute maximum}$ is $ f(2) = 8 $ at $ x = 2 $   and
$\textbf{Absolute minimum}$ is $ f(-2) = -8 $ at $ x = -2 $

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