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Determine the maximum and minimum value of the following function.

f(x) = 2x^{3} – 21x^{2} + 36x – 20

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#### Solution

f(x) = 2x^{3} – 21x^{2} + 36x – 20

∴ f'(x) = 6x^{2} – 42x + 36 and f''(x) = 12x – 42

Consider, f '(x) = 0

∴ 6x^{2} – 42x + 36 = 0

∴ 6(x^{2} – 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.