# Express as a Single Simplified Fraction

1. Sep 14, 2008

### thrill3rnit3

1. The problem statement, all variables and given/known data

sorry I don't know how to type algebraic expressions, but I'll do my best.

express quantity [ f(x+h) - f(x) ] over h as a single simplified fraction

f is not a variable, it's a function

2. Relevant equations

Question # 1. function of x = quantity 1-x over x

3. The attempt at a solution

umm I don't need answers. I just need to know what to do first. Do I substitute the given function of x in the given equation? Do I add the h to the given function? Like say for number 1, do I do (quantity of 1-x over x) + h??

Sorry I'm asking a noob-ish question. I'm not on my best thinking mode right now, especially that things aren't going my way lately...

Thanks :D

Last edited: Sep 14, 2008
2. Sep 14, 2008

### danago

So you are given that $$f(x)=\frac{1-x}{x}$$?

To find f(x+h), all you do is replace any 'x' you see with an 'x+h', so:

$$f(x+h)=\frac{1-(x+h)}{x+h}$$

3. Sep 14, 2008

### praharmitra

do you mean $$f(x) = \frac{1-x}{x}$$ and you need to calculate f(x + h) - f(x).

f(x + h) means that in the given function you must replace 'x' by 'x + h' and NOT adding h to the ENTIRE function

4. Sep 14, 2008

### thrill3rnit3

oh I see, I see. That answers my question. Thanks a lot guys

5. Sep 14, 2008

### thrill3rnit3

so for the first problem, is this correct:

$$\frac{\frac{1-(x+h)}{x+h}-\frac{1-x}{x}}{h}$$

plus further simplification, of course.

6. Sep 14, 2008

### Dick

That's exactly it.

7. Sep 14, 2008

### thrill3rnit3

How about $$\frac{1}{x^{2}}$$ ? Should I replace $$f(x+h)$$ with $$\frac{1}{x^{2}+h}$$ ??

8. Sep 15, 2008

### Dick

No. If f(x)=1/x^2 then f(x+h)=1/(x+h)^2. I thought you had this.

9. Sep 15, 2008

Yep. Thanks.