Verify that the function $f(x)= 2x^3+3x$ satisfies the hypothesis of the mean value theorem on the interval [0,2]. Find all the numbers C that satisfy the conclusion of the Mean Value Theorem.

The mean value of the first derivative of the function on the interval [0,2] is

$[f(2)-f(0)]/(2-0) =(2*2^3+3*2 -0)/2 =(16+6)/2 =11$

the first derivative of the function is $f'(x) =6x^2+3$

Now one must have $f'(x) =11$

$6x^2+3=11$

$6x^2 =8$

$x^2 =4/3$

$x1 =2/\sqrt{(3)}$

$x2 =-2/\sqrt{(3)}$

X1 and x2 are the numbers C that you satisfy the conclusion of the mean value theorem