# Thread: Compute the Difference Quotient

1. ## Compute the Difference Quotient

For the function f(x)=3x^3-34x. Compute the difference quotient f(x+h)-f(x)/h , where h is not equal to 0

2. ## Re: Compute the Difference Quotient

So, what's the problem?

3. ## Re: Compute the Difference Quotient

Hello, spyder12!

I bet you didn't "cube" correctly.

$\text{Given the function: }f(x)\:=\:3x^3-34x$

$\text{Compute the difference quotient: }\:\frac{f(x+h)-f(x)}{h}\,\text{ where }h \ne 0.$

$f(x+h) \;=\;3(x+h)^3 - 34(x+h)$

n . . . . . $=\;3(x^3 + 3x^2h + 3xh^2 + h^3) - 34(x+h)$

n . . . . . $=\;3x^3 + 9x^2h + 9xh^2 + 3h^3 - 34x - 34h$

$f(x+h) - f(x) \;=\;(3x^3 + 9x^2h + 9xh^2 + 3h^3 - 34x - 34h) - (3x^3 - 34x)$

n . . . . . . . . . . $=\;9x^2h + 9xh^2 + 3h^3 - 34h$

$\frac{f(x+h)-f(x)}{h} \;=\;\frac{9x^2h + 9xh^2 + 3h^3 - 34h}{h}$

. . . . . . . . . . . . $=\;\frac{h\,(9x^2 + 9xh + 3h^2 - 34)}{h}$

. . . . . . . . . . . . $=\;9x^2 + 9xh + 3h^2 - 34$

4. ## Re: Compute the Difference Quotient

thank you! you bet right haha