
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.

Just take the derivative and use the quadratic formula. Quick and simple relative to using the mean value theorem.
$\displaystyle \frac{dy}{dx}=3x^26x24=2,3x^26x22=0.$
$\displaystyle \frac{6\pm \sqrt{364(3)(22)}}{6}=\frac{3\pm 5\sqrt{3}}{3}.$
The only one that is in (4,1) is $\displaystyle \frac{35\sqrt{3}}{3}\approx 1.88675.$ This both proves the existence of the value and gives the value explicitly.