Question

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].

Answer 1

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.