What is the relative maximum and minimum of the function f(x) = 2x³ + x² - 11x

Solution:

Given function f(x) = 2x³ + x² - 11x

We know that the function has a minimum when its second derivative function is positive and has a maximum when its second derivative function is negative

f(x) = 2x³ + x² - 11x = 0 to find roots

x(2x² + x - 11) = 0

x = 0 and 2x² + x - 11 = 0

x = -b ± √(b² - 4ac) / 2a

x = -1 ± √(1² - 4(-22)) / 2(2)

x = -1 ± √89 / 4

x = -1 ± 9.4 /4

x = -10.4/4, 8.4/4

x = -2.6, 2.1

f’(x) = 6x² + 2x - 11

f’’(x) = 12x + 2

When x = 0 : f’’(x) = f’’(0)

= 12(0) + 2

= 2 (positive)

Hence , minimum at x = 0

When x = -2.6 : f’’(x) = f’’(-2.6)

= 12(-2.6) + 2

= -31.2 + 2

= -29.2 (negative)

Hence, maximum at x = -2.6

When x = 2.1 : f’’(x) = f’’(2.1)

= 12(2.1) + 2

= 25.2 + 2

= 27.2(positive)

Hence, minimum at x = 2.1

Relative minimum : f(0) = 2(0)³ + (0)² - 11(0) = 0

Relative maximum : f(-2.6) = 2(-2.6) + (-2.6)² - 11(-2.6)

= -5.2 + 6.76 + 28.6

= 30.16

Therefore, the relative maximum is 30.16 and the relative minimum is 0