Advanced Mean Value Theorem - Part 2

Let ${\color{#D61F06}{g\left( x \right)=f\left( x \right){{\left( b-x \right)}^{a}}}}$ , defined on $\left[ a,\ b \right]$ .

$g$ is continuous within the interval $\left[ a,\ b \right]$ , differentiable within the interval $\left( a,\ b \right)$ , with
$\begin{aligned} {g}'\left( x \right) & ={f}'\left( x \right){{\left( b-x \right)}^{a}}+f\left( x \right){{\left[ {{\left( b-x \right)}^{a}} \right]}^{\prime }} \\ & ={f}'\left( x \right){{\left( b-x \right)}^{a}}+f\left( x \right)\left[ a{{\left( b-x \right)}^{a-1}}{{\left( b-x \right)}^{\prime }} \right] \\ & ={{\left( b-x \right)}^{a-1}}\left[ {f}'\left( x \right)\left( b-x \right)-af\left( x \right) \right] \\ \end{aligned}$ Let $h\left( x \right)=\dfrac{f\left( x \right)}{x-a}$ , $x\in \left( a,\ b \right]$ .
Then,
$f\left( x \right)=\left( x-a \right)h\left( x \right)\Rightarrow \underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,\left[ \left( x-a \right)h\left( x \right) \right]=\underset{x\to a}{\mathop{\lim }}\,\left( x-a \right)\cdot \underset{x\to a}{\mathop{\lim }}\,h\left( x \right)=0\cdot 1=0$ Since $f$ is continuous, $f\left( a \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=0$ . Thus, we have $g\left( a \right)=f\left( a \right){{\left( b-a \right)}^{a}}=0$ $g\left( b \right)=f\left( b \right){{\left( b-b \right)}^{a}}=0$ Hence, by Rolle’s theorem , there exists a $\xi\in \left( a,\ b \right)$ such that
${g}'\left( \xi \right)=0\Rightarrow {{\left( b-\xi \right)}^{a-1}}\left[ {f}'\left( \xi \right)\left( b-\xi \right)-af\left( \xi \right) \right]=0\Rightarrow f\left( \xi \right)=\frac{b-\xi }{a}{f}'\left( \xi \right)$

The answer is $\boxed{ \text{Yes, always}}$ .