Homework SourseLet fx22x1 Show that is no value for c such that f3f0 fc30
Let f(x)=2-|2x-1|. Show that is no value for c such that f(3)-f(0) = f\'(c)(3-0). Why does this not contradict the Mean Value Theorem.
Mean Value Theorem:
1. f is continuos on the closed interval [a,b]
2. f is differentiable on the open interval (a,b)
Then there is a number c in (a,b) such that
f\'(c) = f(b)-f(a)/b-a
Mean Value Theorem:
1. f is continuos on the closed interval [a,b]
2. f is differentiable on the open interval (a,b)
Then there is a number c in (a,b) such that
f\'(c) = f(b)-f(a)/b-a
Solution
f(3)-f(0)=-3-1=f\'(c)(3) So f\'(c)=-4 But f\'(x)=-2 for x>1/2 and f\'(x)=2 for x<1/2 So there is no c such that f\'(c)=-4 It conflicts the mean value theorem because [0,3] interval has no differentiability at x=1/2
