I have the following proof of the Mean Value Theorem:
We first need to define a new function. We shall call this function $g$.
$$g(x) = f(x)-(x-a)\frac{f(b)-f(a)}{b-a}$$ for some $x \in [a,b]$.
Although the line above may seem initially daunting, we have in fact simply subtracted a constant multiple of the linear function $(x-a)$ from $f(x)$ and labelled it $g(x)$. The function $g(x)$ is also continuous on $[a,b]$ and differentiable on $(a,b)$. Furthermore we can also see that: $$g(a) = f(a)-\frac{f(b)-f(a)}{b-a}(a-a) = f(a)$$ and $$g(b) = f(b)-\frac{f(b)-f(a)}{b-a}(b-a) = f(a)$$ As we can see, $g(a)=g(b)$, meaning $g$ satisfies Rolle's Theorem. It can be applied and we are now aware that $g'(\xi)=0$ for some $\xi \in (a,b)$. We can now write: $$g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$$ In particular, if we have that $g'(\xi)=0$ then: $$g'(\xi)=f'(\xi)-\frac{f(b)-f(a)}{b-a}=0$$ Naturally this means that: $$f'(\xi)=\frac{f(b)-f(a)}{b-a}$$ We have now obtained what we desired.
My question is: why is it necessary to subtract a constant multiple of the linear function $(x-a)$ from $f(x)$?
Any help in helping me understand this step would be greatly appreciated. Thank you.
It is done so that Rolle's Theorem can be applied. You need $f(a)=f(b)$ in Rolle's Theorem and the new function $g$ satisfies this property. Note that If you take $g(x)=f(x)-c(x-a)$ and you insist on $g(a)=g(b)$ you automatically get $c=\frac {f(b)-f(a)} {b-a}$.