Thread: Using the difference quotient

1. Using the difference quotient

Can somebody please help me out with this?

Let $\displaystyle f(x)=(2/x)-4$, then the expression

$\displaystyle [f(x+h)-f(x)]/h$ can be written in the form

$\displaystyle (A)/(Bx^2+Cxh)$, where A, B, and C are constants.

Find A, B, and C.

Any help would be greatly appreciated. Thanks!

2. Originally Posted by jtowery
Can somebody please help me out with this?

Let $\displaystyle f(x)=(2/x)-4$, then the expression

$\displaystyle [f(x+h)-f(x)]/h$ can be written in the form

$\displaystyle (A)/(Bx^2+Cxh)$, where A, B, and C are constants.

Find A, B, and C.

Any help would be greatly appreciated. Thanks!
$\displaystyle \frac{\frac{2}{x+h}-4-(\frac{2}{x}-4)}{h}$
the 4s cancel
$\displaystyle \frac{\frac{2}{x+h}-\frac{2}{x}}{h}$
divide the h into each part
$\displaystyle \frac{2}{h(x+h)}-\frac{2}{hx}$
make each denominator the same
$\displaystyle \frac{2hx}{h^2x(x+h)}-\frac{2h(x+h)}{h^2x(x+h)}$
combine
$\displaystyle \frac{2hx-2h(x+h)}{h^2x(x+h)}$
1 h on top cancels with 1 h on bottom + factor out 2
$\displaystyle \frac{2(x-x-h)}{hx(x+h)}$
x's on top cancel leaving another h to cancel with the h below
$\displaystyle \frac{-2}{x(x+h)}$
expand denominator
$\displaystyle \frac{-2}{x^2+xh}$
$\displaystyle \therefore a = -2$
$\displaystyle b=1$
$\displaystyle c = 1$

3. nvm?

4. Originally Posted by 1005
nvm
??

$\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{\left(\frac{2}{x+h}-4\right)-\left(\frac{2}{x}-4\right)}{h}=\dots$

Now simplify.

5. Awesome. Thank you both so much