Mark M. answered 01/29/19
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
- f(x) = x2 / (3x-3) is continuous on [2,4] and is differentiable on (2,4). So, by the Mean Value Theorem, there is at least one number, c, in the interval (2,4) so that f'(c) = [f(4)-f(2)] / (4-2).
Find c:
(3c2 - 6c) / (3c-3)2 = 2/9
Simplify to get c2 - 2c - 2 = 0
So, c = 1+√3 (1-√3 is not in the interval (2,4))
For the second problem, the Mean Value Theorem does not apply as the function f(x) = x2 / (2x+2) is not continuous on the interval [-4,0]. (-1 is a dscontinuity)
Mark M.
01/29/19
Elvin R.
Sorry for being rude or sounding like an idiot, but for the first problem, you did the quadratic formula?01/29/19