Find the absolute (global) maximum $ y $ value of the function:$ f(x)=2x^3+0x^2+-24x+5 $On the interval $ [-9, 10] $

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We are asked to find the absolute (global) maximum value of f(x) = 2x^3 + 0x^2 - 24x + 5 on the interval [-9, 10].
First, identify potential extreme points by the standard calculus approach: locate critical points where f'(x) = 0 or undefined, and also evaluate the function at the endpoints of the interval.
Compute the derivative: f'(x) = 6x^2 - 24 = 6(x^2 - 4).
Solve f'(x) = 0: x^2 - 4 = 0, so x = ±2. These are the critical points to check, provided they lie in [-9, 10] (they do).
Next, evaluate the function at the relevant x-values: the endpoints x = -9 and x = 10, and the critical points x = -2 and x = 2.
- At x = -9: f(-9) = 2(-9)^3 - 24(-9) + 5 = 2(-729) + 216 + 5 = -1458 + 216 + 5 = -1237.
- At x = -2: f(-2) = 2(-2)^3 - 24(-2) + 5 = 2(-8) + 48 + 5 = -16 + 48 + 5 = 37.
- At x = 2: f(2) = 2(2)^3 - 24(2) + 5 = 2(8) - 48 + 5 = 16 - 48 + 5 = -27.
- At x = 10: f(10) = 2(10)^3 - 24(10) + 5 = 2(1000) - 240 + 5 = 2000 - 240 + 5 = 1765.
Now compare the values to determine which is largest on the interval. The smallest values among these are at x = -9 and x = 2, and the largest value occurs at x = 10 with f(10) = 1765. Since these are the only candidates (endpoints and critical points) within the interval, the absolute maximum value on [-9, 10] is the largest of { -1237, 37, -27, 1765 }, which is 1765. The process hinges on evaluating all critical points inside the interval and the endpoints, then selecting the maximum among those computed values.
Note: If the derivative had no real zeros inside the interval, we would only compare the endpoint values. Here, the critical points x = ±2 provided additional candidate values to compare against the endpoints.

UCIC 202503 EMTH118 Quiz 10

Find the absolute (global) maximum \( y \) value of the function:\( f(x)=2x^3+0x^2+-24x+5 \)On the interval \( [-9, 10] \)

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