Just take the derivative and use the quadratic formula. Quick and simple relative to using the mean value theorem.
The only one that is in (-4,1) is This both proves the existence of the value and gives the value explicitly.
Prove there exists a number c on (-4,1), such that g ' (c) = -2 Give an explanation for your proof. I used the mean value theorem and found that it indeed does work. But how do I word the explanation.
g(x)= x^3 - 3x^2 - 24x +12
Maybe...
For the continuous function g, the endpoints of the function on the interval (-4,1) form a secant line. For the function g, there exists a point such that the tangent line at that point is parallel to the secant line...
what else should I add...
thanks,
D.