f(x) is cont. and differentiable. If f(-3)=4 and f(7)=2 then

[katie9426]

I'm completely lost on this problem! Any help would be greatly appreciated!

The graph of y=f(x) is continuous on the closed interval [-3,7] and differentiable on the open interval (-3,7) if f(-3)=4 and f(7)=2, then there exists a c, -3<c<7 such that:

. . .f(c)=0, f'(c)=0, f'(c)= 1/5, f'(c)= -1/5, or f'(c)= -5

I dont even know where to start! Thanks for any help!

 

[skeeter]

you should recognize this as an application of the Mean Value Theorem (MVT) ...

the theorem states that if a function is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one x-value "c" in the interval (a,b) such that the average rate of change equals the instantaneous rate of change.

note that the slope of the secant line (also called the average rate of change) on the interval [-3,7] is m = [f(7) - f(-3)]/[7 - (-3)] = (2 - 4)/10 = -1/5

The MVT guarantees the existence of an x-value "c" in the interval (-3,7) such that
the instantaneous rate of change equals the average rate of change, or f'(c) = -1/5.