Discussion
The absolute maximum value of the function \(f(x) = x^{3} - 3x + 2\) in [0,2] is
Answer: E
The absolute maximum is 4.
\(f'(x) = 3x^{2} - 3 = 0\) gives \(x = \pm 1\); only \(x = 1\) lies in [0, 2]. Compare the critical point with the endpoints:
\[ \begin{aligned} f(0) &= 2 \\ f(1) &= 1 - 3 + 2 = 0 \\ f(2) &= 8 - 6 + 2 = 4 \end{aligned} \]
The largest is 4, at the endpoint \(x = 2\). 0 is the absolute minimum, and 2 is the value at the left endpoint. Checking only the critical point misses the maximum, which is at an end of the interval.
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