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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 $(x)$.

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The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:

$\displaystyle\frac{y-f(a)}{x-a}=\frac{f(b)-f(a)}{b-a}$

$\displaystyle\implies g(x)=y=\Big[\frac{f(b)-f(a)}{b-a}\Big](x-a)+f(a)$

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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 $g(b)=f(b)$.

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  • $\begingroup$ You probably meant $g(a)=f(a)$ and $g(b)=f(b)$. $\endgroup$ – Taladris Dec 16 '18 at 14:21
  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Dec 16 '18 at 14:23

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