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