I have these 2 question:
- A function f(x), is any function, so that $L_{a}(x)=L_{b}(x), a \not= b$. Show $\frac{f(b)-f(a)}{b-a}=\frac{f'(b)+f'(a)}{2}$.
- Assume that $a \not= b$ and that $\frac{f(b)-f(a)}{b-a}=\frac{f'(b)+f'(a)}{2}$, prove or disprove $L_{a}(x)=L_{b}(x)$.
For 1. I find that $\frac{f(b)-f(a)}{b-a}=\frac{f'(b)+f'(a)}{2}$ this is asking when the average rate of change is equal to the instantaneous rate of change, of a and b. But, that is all I find. How do I show this equation is true using this information?
For 2. I know $L_{a}(x)=f(a)$ and $L_{b}(x)=f(b)$. They are differentiable. at x=a, x=b, for f(x). I think, the linear approximation is a linear equation. And, a linear equation is a polynomial, so if $L_{a}(x)=L_{b}(x)$, then every term in the linear equation is equal.
Thank you for the help.
So it I understand well $L_a(x)$ being the linear approximation of $f$ at $a$, we have
$$L_a(x) = f^\prime(a)(x-a) +f(a)$$
and in a similar way $$L_b(x) = f^\prime(b)(x-b) +f(b).$$
If those two coincide, we have in particular $$\begin{cases} L_a(b)&=f(b)&=f^\prime(a)(b-a) +f(a)\\ L_b(a)&=f(a)&=f^\prime(b)(a-b) +f(b) \end{cases}$$
From which we get the requested equality by substracting the two previous equations.
The converse is not true. For a counterexample, take $f(x)=x^2$ and $a=-b=1$. We have
$$\frac{f(b)-f(a)}{a-b} = \frac{f^\prime(a)+f^\prime(b)}{2}=0$$ however the linear approximations are different.