# More Difference Quotient

• Dec 23rd 2008, 11:08 AM
magentarita
More Difference Quotient
Find the difference quotient for the given function.

y = x + (1/x)
• Dec 23rd 2008, 12:10 PM
Pn0yS0ld13r
Quote:

Originally Posted by magentarita
Find the difference quotient for the given function.

y = x + (1/x)

let $\displaystyle f(x) = x+\dfrac{1}{x}$. Now just plug in the following equation:

Difference quotient: $\displaystyle \dfrac{f(x+h) - f(x)}{h}$
• Dec 23rd 2008, 08:57 PM
magentarita
no........
Quote:

Originally Posted by Pn0yS0ld13r
let $\displaystyle f(x) = x+\dfrac{1}{x}$. Now just plug in the following equation:

Difference quotient: $\displaystyle \dfrac{f(x+h) - f(x)}{h}$

I'm sorry but reaching the answer involves more than what you suggested.
• Dec 23rd 2008, 09:23 PM
Chris L T521
Quote:

Originally Posted by magentarita
Find the difference quotient for the given function.

y = x + (1/x)

Using the definition of the difference quotient, we have $\displaystyle \frac{x+h+\displaystyle\frac{1}{x+h}-\left[x+\displaystyle\frac{1}{x}\right]}{h}=\frac{h+\displaystyle\frac{1}{x+h}-\frac{1}{x}}{h}$

The goal here is to get the h term in the denominator to disappear.

First, we need to combine the terms in the numerator.

The common denominator is $\displaystyle x(x+h)$

Thus, the difference quotient becomes $\displaystyle \frac{\displaystyle\frac{hx(x+h)}{x(x+h)}+\frac{x} {x(x+h)}-\frac{x+h}{x(x+h)}}{h}=\frac{\displaystyle\frac{hx ^2+h^2x-h}{x(x+h)}}{h}=\frac{h\left(x^2+hx-1\right)}{h\left[x(x+h)\right]}$ $\displaystyle =\color{red}\boxed{\frac{x^2+hx-1}{x(x+h)}}$

Does this make sense?

By the way, to add on to what I told you in the other thread, when simplifying the difference quotient, the objective is to get the h term in the denominator to disappear (i.e. cancel out with something in the numerator of the difference quotient).
• Dec 28th 2008, 07:14 AM
magentarita
a lot...
Quote:

Originally Posted by Chris L T521
Using the definition of the difference quotient, we have $\displaystyle \frac{x+h+\displaystyle\frac{1}{x+h}-\left[x+\displaystyle\frac{1}{x}\right]}{h}=\frac{h+\displaystyle\frac{1}{x+h}-\frac{1}{x}}{h}$

The goal here is to get the h term in the denominator to disappear.

First, we need to combine the terms in the numerator.

The common denominator is $\displaystyle x(x+h)$

Thus, the difference quotient becomes $\displaystyle \frac{\displaystyle\frac{hx(x+h)}{x(x+h)}+\frac{x} {x(x+h)}-\frac{x+h}{x(x+h)}}{h}=\frac{\displaystyle\frac{hx ^2+h^2x-h}{x(x+h)}}{h}=\frac{h\left(x^2+hx-1\right)}{h\left[x(x+h)\right]}$ $\displaystyle =\color{red}\boxed{\frac{x^2+hx-1}{x(x+h)}}$

Does this make sense?

By the way, to add on to what I told you in the other thread, when simplifying the difference quotient, the objective is to get the h term in the denominator to disappear (i.e. cancel out with something in the numerator of the difference quotient).

That's a lot of math, mate.