# Math Help - Functions Question 2

1. ## Functions Question 2

Given $f(x)=\frac{1}{x}$.Find $\frac{f(x+h)+f(x)}{h}$ where h is not equal to zero

Attempt:
$\frac{\frac{h}{x+h*x)}}{h}$
How to simplify (please provide steps)

Thank you

2. Originally Posted by mj.alawami
Given $f(x)=\frac{1}{x}$.Find $\frac{f(x+h)+f(x)}{h}$ where h is not equal to zero

Attempt:
$\frac{\frac{h}{x+h*x)}}{h}$
How to simplify (please provide steps)

Thank you
$\frac{\frac{1}{x+h}+\frac{1}{x}}{h}$

$\frac{2x+h}{x^2+xh}\cdot \frac{1}{h}$

$
\frac{2x+h}{x^2h+xh^2}
$

3. Originally Posted by mj.alawami
Given $f(x)=\frac{1}{x}$.Find $\frac{f(x+h)+f(x)}{h}$ where h is not equal to zero
you sure it's not $\frac{f(x+h)-f(x)}{h}$

4. Originally Posted by skeeter
you sure it's not $\frac{f(x+h)-f(x)}{h}$
yes you are correct and i am wrong? So how is it done ?

5. Originally Posted by mj.alawami
Given $f(x)=\frac{1}{x}$.Find $\frac{f(x+h)+f(x)}{h}$ where h is not equal to zero

Attempt:
$\frac{\frac{h}{x+h*x)}}{h}$
How to simplify (please provide steps)

Thank you
$f(x) = \frac{1}{x}$

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

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

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

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

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

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

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

6. Originally Posted by mj.alawami
So how is it done ?
To learn, in general, how to evaluate a function at an expression (rather than a number), try here.

Then try working this exercise in pieces, rather than all at one. First write down the formula for f(x). Then find the expression for f(x + h). Then subtract the polynomial for f(x) from the polynomial for f(x + h). Then divide by h, cancelling if you can.