To find the maximum and minimum values of the function f(x) = 2x^3 - 4x + 3, we need to find the first derivative.

f(x) = 2x^3 - 4x + 3

f’(x) = 6x^2 - 4

Equate this to zero and solve for x

6x^2 - 4 = 0

=>...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

To find the maximum and minimum values of the function f(x) = 2x^3 - 4x + 3, we need to find the first derivative.

f(x) = 2x^3 - 4x + 3

f’(x) = 6x^2 - 4

Equate this to zero and solve for x

6x^2 - 4 = 0

=> x ^2 = 4/6

=> x = +sqrt (2/3) and x = -sqrt (2/3)

To determine whether the values are minimum or maximum, we find the second derivative

f’’(x) = 12x.

As f’’(x) is positive at x = +sqrt (2/3), we have a minimum value here.

And as f’’(x) is negative at x = -sqrt (2/3), we have a maximum value here.

For x = +sqrt (2/3), f(x) = 2*(2/3)*sqrt (2/3) - 4*sqrt (2/3) + 3

For x = -sqrt (2/3), f(x) = -2*2/3*sqrt (2/3) + 4*sqrt (2/3) +3

**The maximum value of the function is (8/3)*sqrt(2/3)+ 3 at x = -sqrt(2/3)**

**And the minimum value of the function is -(8/3)*sqrt(2/3) + 3 at x = +sqrt(2/3).**