Given that $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, it is to be proved that there exists an $\xi \in (a,b)$ such that $$f'(\xi)=\dfrac{2}{a-\xi}\cdot (f(\xi)-f(b)).$$
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Let $$F(x)=(x-a)^2(f(x)-f(b)),x\in[a,b].$$ Then $F(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $$F(a)=F(b)=0.$$ By Rolle's MVT, we know there exists $\xi\in(a,b)$ such that $F'(\xi)=0.$ It is easy to see that $$F'(x)=2(x-a)(f(x)-f(b))+(x-a)^2f'(x).$$ So $$F'(\xi)=0\iff f'(\xi)=\dfrac{2}{a-\xi}\cdot (f(\xi-f(b)).$$
Riemann
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