Math Help - Another proof question

1. Another proof question

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.

2. Just take the derivative and use the quadratic formula. Quick and simple relative to using the mean value theorem.

$\frac{dy}{dx}=3x^2-6x-24=-2,3x^2-6x-22=0.$

$\frac{6\pm \sqrt{36-4(3)(-22)}}{6}=\frac{3\pm 5\sqrt{3}}{3}.$

The only one that is in (-4,1) is $\frac{3-5\sqrt{3}}{3}\approx -1.88675.$ This both proves the existence of the value and gives the value explicitly.