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Question
Determine the minimum value of the 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
∴ x2 – 7x + 6 = 0
∴ (x – 6)(x – 1) = 0
∴ x – 1 = 0 or x – 6 = 0
∴x = 6 or x = 1
For x = 6,
f''(6) = 12(6) – 42
= 72 – 42
= 30 > 0
∴ f(x) minimum value at x = 6.
f''(1) = 12 – 42
= –30 < 0
∴ f(x) maximum value at x = 1.
f''(6) = 2(6)3 – 21(6)2 + 36(6) – 20
= 432 – 756 + 216 – 20
= –128
∴ The function f(x) has a minimum value of –128 at x = 6.
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