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Question
Find the local maximum and local minimum value of f(x) = x3 − 3x2 − 24x + 5
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Solution
f(x) = x3 − 3x2 − 24x + 5
∴ f′(x) = 3x2 – 6x – 24
∴ f''(x) = 6x − 6
Consider, f′(x) = 0
∴ 3x2 – 6x – 24 = 0
∴ 3(x2 – 2x – 8) = 0
∴ x2 – 2x – 8 = 0
∴ (x + 2)(x – 4) = 0
∴ x + 2 = 0 or x – 4 = 0
∴ x = – 2 or x = 4
For x = – 2,
f ''(– 2) = 6(– 2) − 6
= −18 < 0
∴ f(x) is maximum at x = – 2.
∴ Maximum value = f(–2) = (–2)3 − 3(–2)2 – 24(–2) + 5
= – 8 – 12 + 48 + 5
= 33
For x = 4,
f''(4) = 6(4) − 6
= 18 > 0
∴ f(x) is minimum at x = 4.
∴ Minimum value = f(4) = (4)3 − 3 (4)2 – 24 (4) + 5
= 64 − 48 – 96 + 5
= −75
∴ Local maximum of f(x) is 33 when x = – 2 and Local minimum of f(x) is −75 when x = 4.
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