1. ## Slope and Gradient Proof

Hello,

I have to show that for some constant m:
$\displaystyle \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}=m$

For all $\displaystyle x_{1}$≠$\displaystyle x_{2}$, show that f(x)=mx+b, where b is some constant. The hint given is to fix $\displaystyle x_{1}$ and take x=x$\displaystyle x_{2}$; then solve for f(x).

I figured that I should change it to point slope form like this (referring to the hint given):

But I'm not sure if this step is correct or not and how I can turn it into the slope-intercept form? Please help.

2. Originally Posted by Kataangel
Hello,

I have to show that for some constant m:
$\displaystyle \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}=m$

For all $\displaystyle x_{1}$≠$\displaystyle x_{2}$, show that f(x)=mx+b, where b is some constant. The hint given is to fix $\displaystyle x_{1}$ and take x=x$\displaystyle x_{2}$; then solve for f(x).

I figured that I should change it to point slope form like this (referring to the hint given):

But I'm not sure if this step is correct or not and how I can turn it into the slope-intercept form? Please help.
Yes, that step is perfectly good (it uses the "distributive" law) and it is in "slope-intercept form". Just think of it as $\displaystyle y= mx+ (f(x_1)- mx_1)$. m is the slope and $\displaystyle f(x_1)- mx_1$ is the intercept.

3. Originally Posted by HallsofIvy
Yes, that step is perfectly good (it uses the "distributive" law) and it is in "slope-intercept form". Just think of it as $\displaystyle y= mx+ (f(x_1)- mx_1)$. m is the slope and $\displaystyle f(x_1)- mx_1$ is the intercept.
How would $\displaystyle f(x_1)- mx_1$ be a constant though since it consists of 2 variables?

### PROOF OF SLOPE AND GRADIENT

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