The MVT.

VonNemo19 · #1 $Old$ June 27th, 2009, 06:17 PM

Right now I'm studying the proof of the mean value theorem and ther's something about it that I am having trouble understanding.

I'm going to attemp to explain where I'm having trouble with the proof.

Refer to the attached figure during this discussion.

In the proof, the first thing that they do is say that the equation to the secant line is

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

I understand that $\left[\frac{f(b)-f(a)}{b-a}\right]$ is the slope of the secant line, but I can't understand the rest. My mind wants to say that $(x-a)$ is x (in the slope intercept form of a line) and $-f(a)=b$ .

Could someone justify this as best they can?

Shyam · #2 $Old$ June 27th, 2009, 06:25 PM

Quote:

Originally Posted by VonNemo19 $View Post$

Right now I'm studying the proof of the mean value theorem and ther's something about it that I am having trouble understanding.

I'm going to attemp to explain where I'm having trouble with the proof.

Refer to the attached figure during this discussion.

In the proof, the first thing that they do is say that the equation to the secant line is

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

I understand that $\left[\frac{f(b)-f(a)}{b-a}\right]$ is the slope of the secant line, but I can't understand the rest. My mind wants to say that $(x-a)$ is x (in the slope intercept form of a line) and $-f(a)=b$ .

Could someone justify this as best they can?

There is point-slope form of a line

$y - y_1=m(x -x_1)$ where m is slope and $(x_1, \;y_1)$ is a point on the line.

$y =m(x -x_1)+y_1$

here slope is $m=\frac{f(b)-f(a)}{b-a}$ and point $(x_1, \;y_1) =(a, f(a))$

therefore, the equation becomes,

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

VonNemo19 · #3 $Old$ June 27th, 2009, 06:30 PM

Quote:

Originally Posted by Shyam $View Post$

and $(x_1, \;y_1) =(a, f(a))$

therefore, the equation becomes,

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

I can't believe that I didn't see that. Thanks man.