# Is inverse use of mean value theorem right?

If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then

for any $x\in (a,b)$, exists $y, z \in [a,b]$, such that

$f '(x)=\dfrac{f(z)-f(y)}{z-y}$

Is this right?

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No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.