Question

Determine the absolute extrema of the function 𝑓(𝑥)=𝑥3−18𝑥2 96𝑥 , on the interval [1,10]

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Seamus

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Answer:
absolute minimum is at f(10) = -704
absolute maximum is at f(1) = 79
Solution:
f(x) = x³ - 18x² + 96x
f'(x) = 3x² - 36x + 96
then solve for x where f'(x) = 0
f'(x) = 3x² - 36x + 96
0 = 3x² - 36x + 96
0 = 3(x² - 12x + 32)
0 = x² - 12x + 32
0 = (x - 4)(x - 8)
x = 4
x = 8
Since 4 and 8 is in the interval [1,10] we can solve for f(x) using 1,4,8, and 10
f(x) = x³ - 18x² + 96x
f(1) = 1³ - 18(1)² + 96(1) = 79
f(4) = 4³ - 18(4)² + 96(4) = -128
f(8) = 8³ - 18(8)² + 96(8) = -544
f(10) = 10³ - 18(10)² + 96(10) = -704
making the value at f(10) the absolute minimum and the value at f(1) the absolute maximum

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