f(x) = {cos x, 0

Question

f(x) = {cos x, 0<=x<pi/2 and {sin x, pi/2<=x<=pi. Both on the interval [0,pi]. State whether or not the function satisfies the hypotheses of the Mean Value Theorem on the given interval, and if it does, find each value of c in the interval (a,b) that satisfies the equation f'(c)= (f(b)-f(a))/(b-a).

If someone can do me the favor and explain Mean Value Theorem to me, that would be fantastic.

anonymous · Answer

Mean Value Theorem:  Let f be a continuous function on a compact interval [a,b] that is differentiable at every point in the interior.  Then there exists a point c in the interior where$$f'(c)=\frac{f(b)-f(a)}{b-a}$$So if the function is $$f(x)=\left[\begin{matrix}\cos(x) & 0 \le x < \frac{\pi}{2} \ \sin(x) & \frac{\pi}{2} \le x \le \pi \end{matrix}\right]$$on$$[0, \pi]$$we first check to see if the interval is compact.  And, since it is indeed closed and bounded we know that it is compact.

We can also see that the function is differentiable at every point in the interior, since both cos(x) and sin(x) are differentiable at every point.

However, the function is not continuous.  As$$\lim_{x \rightarrow \frac{\pi}{2}^{-}}\cos(x)\rightarrow 0$$and$$\lim_{x \rightarrow \frac{\pi}{2}^{+}}\sin(x) \rightarrow 1$$