The mean values theorem says that there exists a $c∈(u,v)$ such that $$f(v)-f(u)=f′(c)(v-u)$$ My question is: Can we inverse this situation, i.e., given the function $f$ and the real $c$, can we find $u,v$ such that $$f(v)-f(u)=f′(c)(v-u)$$ holds true?
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Nope! Have a look at $x^3$ at the origin.
Graphically, you are drawing a tangent line at $c$ and then sliding it up and down and seeing if you can make it hit the curve at two points $u,v$, which allows you to build other examples too! Have a think to see if you can come up with some restrictions that make your result hold.
answered Apr 19, 2013 at 15:58