What is the maximum and minimum value of the function f(x) = 2x^3 - 4x + 3

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).