# Equation of secant line in mean value theorem proof

I'm going through a proof for the mean value theorem.

We have a function $$f(x)$$ continuous on $$[a, b]$$ and differentiable on $$(a, b)$$.

Then we define a function $$g(x)$$ to be the secant line passing through $$(a, f(a))$$ and $$(b, f(b))$$.

The slope of said secant is:

$$m=\frac{f(b)-f(a)}{b-a}$$

That is clear. Now the proof I'm following defines $$g(x)$$ like so:

$$g(x) = \left[ \frac{f(b)-f(a)}{b-a} \right](x-a)+f(a)$$

What confuses me: why is the coefficient defined to be $$(x-a)$$ and not simply $$(x)$$.

The equation of the straight line passing through $$(a,f(a)),(b,f(b))$$ is given as:
$$\displaystyle\frac{y-f(a)}{x-a}=\frac{f(b)-f(a)}{b-a}$$
$$\displaystyle\implies g(x)=y=\Big[\frac{f(b)-f(a)}{b-a}\Big](x-a)+f(a)$$
First of all, it is not correct to assert that $$g(x)$$ is the secant line passing through $$\bigl(a,f(a)\bigr)$$ and $$\bigl(b,f(b)\bigr)$$. It is the function whose graph is the line segment uniting those two points.
Now, concerning your question: it is so that $$g(a)=f(a)$$ and $$g(b)=f(b)$$.
• You probably meant $g(a)=f(a)$ and $g(b)=f(b)$. – Taladris Dec 16 '18 at 14:21