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Question
Find the maximum and minimum of the following functions : f(x) = x3 – 9x2 + 24x
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Solution
f(x) = x3 – 9x2 + 24x
∴ f'(x) = `d/dx(x^3 - 9x^2 + 24x)`
= 3x2 – 9 x 2x + 24 x 1
= 3x2 – 18x + 24
and
f''(x) = `d/dx(3x^2 - 18x + 24)`
= 3 x 2x – 18 x 1 + 0
= 6x – 18
f''(x) = 0 gives 3x2 – 18x + 24 = 0
∴ x2 – 6 x + 8 = 0
∴ (x – 2)(x – 4) = 0
∴ the roots of f'(x) = 0 are x1 = 2 and x2 = 4.
(a) f"(2)
= 6(2) – 18
= – 6 < 0
∴ by the second derivaative test, f has maximum at x = 2 and maximum value of at x = 2
= f(2)
= (2)3 – 9(2)2 + 24(2)
= 8 – 36 + 48
= 20
(b) f"(4) = 6(4) – 18 = 6 > 0
∴ by the second derivaative test, f has maximum at x = 4 and maximum value of at x = 4
= f(4)
= (4)3 – 9(4)2 + 24(4)
= 64 – 144 + 96
= 16.
Hence, the function f has maximum value 20 at x = 2 and minimum value 16 at x = 4.
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