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Question
Determine the maximum and minimum value of the following function.
f(x) = 2x3 – 21x2 + 36x – 20
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Solution
f(x) = 2x3 – 21x2 + 36x – 20
∴ f'(x) = 6x2 – 42x + 36 and f''(x) = 12x – 42
Consider, f '(x) = 0
∴ 6x2 – 42x + 36 = 0
∴ 6(x2 – 7x + 6) = 0
∴ 6(x – 1)(x - 6) = 0
∴ (x – 1)(x – 6) = 0
∴ x – 1 = 0 or x – 6 = 0
∴ x = 1 or x = 6
For x = 1,
f''(1) = 12(1) – 42 = 12 – 42 = – 30 < 0
∴ f(x) attains maximum value at x = 1.
∴ Maximum value = f(1)
= 2(1)3 – 21(1)2 + 36(1) – 20
= 2 – 21 + 36 – 20
= – 19 – 20 + 36
= – 39 + 36
= – 3
∴ The function f(x) has maximum value – 3 at x = 1.
For x = 6,
f''(6) = 12(6) – 42 = 72 – 42 = 30 > 0
∴ f(x) attains minimum value at x = 6.
∴ Minimum value = f(6)
= 2(6)3 – 21(6)2 + 36(6) – 20
= 432 – 756 + 216 – 20
= – 128
∴ The function f(x) has minimum value – 128 at x = 6.
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