1. ## Function Review

Hello, I'm trying to do a review of functions and I'm having difficulty on this one:

$\displaystyle f(x) = \frac 1x$

The problem is

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

I've tried a few possible ways, getting the denominators the same in the numerator. No luck. Probably missing something simple.

Can someone point me in the right direction?

2. ## check this out

f(x)=1/x so
x=1/f(x)
denominator = x-a =(1/[f(x)] - 1/[(f(a)])
=[f(a)-f(x)]/f(a)f(x)
=-[f(x)-f(a)]/f(x)f(a)
where f(x)-f(a)=numerator

3. Hello Impalord
Originally Posted by Impalord
Hello, I'm trying to do a review of functions and I'm having difficulty on this one:

$\displaystyle f(x) = \frac 1x$

The problem is

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

I've tried a few possible ways, getting the denominators the same in the numerator. No luck. Probably missing something simple.

Can someone point me in the right direction?
You don't say what it is you want to do with $\displaystyle \frac {f(x) - f(a)}{x-a}$, but if we replace $\displaystyle f(x)$ with $\displaystyle \frac1x$ and $\displaystyle f(a)$ with $\displaystyle \frac1a$, we get:

$\displaystyle \frac {f(x) - f(a)}{x-a}=\frac{\Big(\dfrac1x-\dfrac1a\Big)}{x-a}=\frac{\Big(\dfrac{a-x}{ax}\Big)}{x-a}=\frac{-\Big(\dfrac{x-a}{ax}\Big)}{x-a}=-\frac{1}{ax}$

Does that help?

Hello ImpalordYou don't say what it is you want to do with $\displaystyle \frac {f(x) - f(a)}{x-a}$, but if we replace $\displaystyle f(x)$ with $\displaystyle \frac1x$ and $\displaystyle f(a)$ with $\displaystyle \frac1a$, we get:

$\displaystyle \frac {f(x) - f(a)}{x-a}=\frac{\Big(\dfrac1x-\dfrac1a\Big)}{x-a}=\frac{\Big(\dfrac{a-x}{ax}\Big)}{x-a}=\frac{-\Big(\dfrac{x-a}{ax}\Big)}{x-a}=-\frac{1}{ax}$

Does that help?

$\displaystyle \frac {a-x}{ax} into \frac {-(x-a)}{ax}$