Show that the Mean Value Theorem does not apply to f(x)=x−2 on (−1,1). Is this a contradiction to the Mean Value Theorem?
My solution so far:
f(1)−f(−1)=f′(c)(1−(−1))⇔1−1=2f′(c)⇔f′(c)=0
Since f′(x)=−2x−3⇒f′(c)=−2c−3
f′(c)=−2c−3=0.
This is not satisfied by any value c so I think I've now got the first part of the task. However, I'm not quite sure how do I figure if this contradicts with the Mean Value Theorem. I know that the MVT tells us that if f is continuous on [a,b] and differentiable on (a,b), there is a point c∈(a,b) such that f(b)−f(a)=f′(c)(b−a) but I don't know how to apply it here.
Answer
The mean-value theorem only applies for continous functions. But x−2=1x2 is not defined at x=0 , the singularity is not even removeable.
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