Give an example which demonstrates that continuity is a necessary condition for the mean value theorem:
I thought in this function:
$$g(x) = \begin{cases} x + 1 & x < 1 \\[4pt] x - 1 & 1 \leq x\\ \end{cases} $$
Clearly $g(x)$ is not continuos in $1$, i try to find the derivative of this function but i stuck some help please for prove that for this function can't use the mean value theorem.
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$\begingroup$Consider your function $g$ on the interval $[0,1]$. Then $g$ is indeed differentiable in $(0,1)$ (with $g'(x)=1$ for all $x\in(0,1)$. The claim of the mean value theorem would say that there exists $c\in(0,1)$ with $g'(c)=\frac{g(1)-g(0)}{1-0}$, i.e. $1=-1$.
$\endgroup$ $\begingroup$For your function, $g(0)=g(2)=1$ so if the mean value theorem applied there would be a point beteen 0 and 2 where the derivative was 0. But the derivative is equal to 1 with left and right limits both equal to 1 at $x=1$. Hence the mean value theorem fails for this discontinuous function.
$\endgroup$ 2 $\begingroup$Mean value theorem in its usual formulation applies to differentiable functions. Which should be at least continuous. If a function is not continuous, then there is no tangent in some points (also, it would be strange to ask fo a derivative in jump point, since there g(x + Δx)- g(x) ↛ 0 as Δx→0)
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