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Questions
Find the absolute maximum and minimum values of a function f given by f(x) = 2x3 − 15x2 + 36x + 1 on the interval [1, 5].
Find the absolute maximum and absolute minimum of function f(x) = 2x3 − 15x2 + 36x + 1 on [1, 5].
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Solution
Given: f(x) = 2x3 − 15x2 + 36x + 1
Differentiate w. r. t. x,
f'(x) = 6x2 − 30x + 36
Put f'(x) = 0
6(x2 − 5x + 6) = 0
x2 − 5x + 6 = 0
x2 − 3x − 2x + 6 = 0
x(x − 3) −2(x − 3) = 0
(x − 3)(x − 2) = 0
then, x =3 or 2
x = 1, 2, 3, 5
Now f(1) = 2(1)3 – 15(1)2 + 36(1) + 1
= 2 – 15 + 36 + 1
= 24
f(2) = 2(2)3 – 15(2)2 + 36(2) + 1
= 16 – 60 + 72 + 1
= 29
f(3) = 2(3)3 – 15(3)2 + 36(3) + 1
= 54 – 135 + 108 + 1
= 28
f(5) = 2(5)3 – 15(5)2 + 36(5) + 1
= 250 – 375 + 180 + 1
= 56
Hence, the absolute maximum value when x = 5 is 56, and the absolute minimum value when x = 1 is 24.
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