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:


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:


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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.