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The bonus question:

Let $f(x)$ be defined for $a \leq x \leq b$. Assuming appropriate properties of continuity and derivability, prove for $a < x < b$ that $ \frac{\frac{f(x)-f(a)}{x-a} - \frac{f(b)-f(a)}{b-a}}{x-b} = \tfrac{1}{2}f^{\prime \prime}(\beta) $ where $\beta$ is some number between $a$ and $b$.

I am thinking that its just the Mean Value Theorem.

$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

I have rewritten the problem as this:

$\frac{f^{\prime}(c)-f^{\prime}(d)}{c-d}$ for $c$ and $d$ within $(a,b)$.

Then I get lost, I don't see where the contant $1/2$ comes into play and how to have the denomiator be $x-b$.

1 Answers 1

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Putting these pieces together should give you the result

$f(x)=f(a)+f'(a)(x-a)+f''(\beta)\frac{(x-a)^2}{2}$

$g(x)=\frac{f(x)-f(a)}{x-a}, g'(x)=\frac{f'(x)}{x-a}- \frac{f(x)-f(a)}{(x-a)^2}$

$ \frac{g(x) - g(b)}{x-b}=g'(c)$

Edit: The first formula is from the Taylor expansion of $f$ You can also derive it using twice rolle on $ h(t)= f(t)-f(a)-f'(a)(t-a)- \frac{(t-a)^2}{(x-a)^2}(f(x)-f(a)-f'(a)(x-a))$ first rolle $h(a)=h(x)=0 \Rightarrow h'(x_1)=0 x_1\in (a,x)$

second rolle on $h'(a)=h'(x_1)=0 \Rightarrow h''(\beta)=0 \beta \in (0,x_1)$

And $h''(\beta)=0$ is what you want

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