According to Mean-Value Theorem (MVT) in Calculus, "There is a point $c$, such that $a < c < b$, at which the tangent line is parallel to the secant line through $(a,y(a))$ and $(a, y(b))$". I have these 2 graphs and which do not conform to MVT. I have my reasons here why they do not, I want to know if I am correct.

asymptotic discontinuity jump discontinuity

Red lines are $Secant-Lines$. It can be easily seen that graph-1 has asymptotic discontinuity and graph-2 has jump discontinuity. Secant lines are red and there can be drawn not one tangent line that is parallel to the secant line in both graphs. Hence, MVT is not applicable .

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    $\begingroup$ You are correct. Simply put, the function has to in fact be differentiable in the open interval between the points, but here it is not even continuous : in one the function is unbounded at a point, and the other involves a jump discontinuity. $\endgroup$ – астон вілла тереса лисбон Aug 29 '18 at 14:34

The MVT requires the assumption that $f$ is differentiable in $(a,b)$. Your examples show functions with infinities or jump discontinuities. These are not even continuous, so clearly not differentiable.

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