Find the absolute minimum and absolute maximum values of
𝑓
(
𝑥
)
=
𝑥
3
−
3
𝑥
+
2
on the closed interval
[
−
2
,
3
]
.

Question

Find the absolute minimum and absolute maximum values of 
𝑓
(
𝑥
)
=
𝑥
3
−
3
𝑥
+
2
 on the closed interval 
[
−
2
,
3
]
.

BlackTom AI · Accepted Answer

To determine the absolute extremes of f(x) = x^3 - 3x + 2 on the closed interval [-2, 3], I’ll examine critical points and the endpoints.
First, compute the derivative: f'(x) = 3x^2 - 3 = 3(x^2 - 1). Setting f'(x) = 0 gives x^2 = 1, so x = ±1 are critical points inside the interval.
Next, evaluate the function at the candidate points: endpoints and critical points.
- At x = -2 (left endpoint): f(-2) = (-2)^3 - 3(-2) + 2 = -8 + 6 + 2 = 0.
- At x = 3 (right endpoint): f(3) = 3^3 - 3(3) + 2 = 27 - 9 + 2 = 20.
- At x = -1 (critical point): f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4.
- At x = 1 (critical point): f(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.
Now compare these values: 0, 20, 4, and 0. The smallest value is 0, occurring at x = -2 and x = 1, so that is the absolute minimum. The largest value is 20, occurring at x = 3, so that is the absolute maximum.
Therefore, the absolute minimum is 0 and the absolute maximum is 20, which matches the provided answer of "0 and 20".

Find the absolute minimum and absolute maximum values of 𝑓 ( 𝑥 ) = 𝑥 3 − 3 𝑥 + 2 on the closed interval [ − 2 , 3 ] .单项选择题

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