Find the absolute maximum and minimum values of the function $ f(x) = (x - 1)^2 + 3 $ in the interval $ [-3, 1] $.

Q: Find the absolute maximum and minimum values of the function $ f(x) = (x - 1)^2 + 3 $ in the interval $ [-3, 1] $.

Explanation: We have $ f(x) = (x - 1)^2 + 3 $ \[ f'(x) = 2(x - 1) \Rightarrow f'(x) = 0 \Rightarrow x = 1 \] Now evaluate $ f $ at endpoints and critical point: \[ f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19 \\ f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3 \] $\textbf{Absolute maximum}$ is $ 19 $ at $ x = -3 $ and $\textbf{Absolute minimum}$ is $ 3 $ at $ x = 1 $

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Explanation: We have $ f(x) = (x - 1)^2 + 3 $

$$
f'(x) = 2(x - 1) \Rightarrow f'(x) = 0 \Rightarrow x = 1
$$

Now evaluate $ f $ at endpoints and critical point:
$$
f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19 \
f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3
$$

$\textbf{Absolute maximum}$ is $ 19 $ at $ x = -3 $  and  $\textbf{Absolute minimum}$ is $ 3 $ at $ x = 1 $

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