Given f(x)= -1/x, find all c in the interval [-3, -1/2] that satisfies the Mean Value Theorem. How do you solve this problem and which is the right answer choice and why? a) c= -√3/2 b) + & - √3/2 c) The Mean Value Theorem doesn't apply because f is not continuous at x=0 d) The Mean Value Theorem doesn't apply because f(-1/2) doesn't equal f(-3) e) none of the above

First we need the slope of the secant line through the points at x = -3 and x = -1/2. The two points are (-3,1/3) and (-1/2,2).

This slope is (2-1/3)/(-1/2-(-3)) = (5/3)/(5/2) = 2/3.

So we seek all c in [-3,-1/2] where f'(x) = 1/x^{2} is 2/3.

1/x^{2 }= 2/3

x^{2 }= 3/2

x = -√(3/2) which is the only one in our interval. This is choice a.