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+1 vote
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in Applications of Differential Calculus by (48.7k points)
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Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x3 – 12x + 1, [-3, 5]

1 Answer

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Best answer

f(x) = x3 – 12x + 1 

f'(x) = 3x2 – 12 

f'(x) = 0 ⇒ 3x2 – 12 = 0 

3x2 = 12 ⇒ x2 = 4 ⇒ x = ±2 

The x values are -3, -2, 2, 5 

f(x) (at x = -3) = (-3)3 – 12 (-3) + 1 = -27 + 36 + 1 = 10

f(x) (at x = -2) = (-2)3 – (12)(-2) + 1 = -8 + 24 + 1 = 17 

f(x) (at x = 2) = 23 – 12 (2) + 1 = 8 – 24 + 1 = -15 

f(x) (at x = 5) = 53 – 12 (5) + 1 = 125 – 60 + 1 = 66 

From the above four values 10,17,-15 and 66, we see that absolute maximum is 66 and the absolute minimum is -15.

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