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