Question

if f(x) = (3x)/(-x +2) find (f(x+h) - f(x))/h and simplify completely

Answer 1

\[f(x)=\frac{3x}{2-x}~~\Rightarrow~~f(x+h)=\frac{3(x+h)}{2-(x+h)}\]
So you have
\[\frac{f(x+h)-f(x)}{h}=\frac{1}{h}\left(\frac{3(x+h)}{2-(x+h)}-\frac{3x}{2-x}\right)\]

Answer 2

do i just distribute everything out now?

Answer 3

Finding a common denominator first might help. You should combine the fractions.

Answer 4

um ok

Answer 5

Looks like @KeithAfasCalcLover is in the process of typing it all out. I'll leave it to him.

Answer 6

can i cross out the (-2x+2h)?

Answer 7

and (2-x)?

Answer 8

yes no?

Answer 9

\[f(x)=\frac{3x}{2-x}\]
\[\frac{f(x+h)-f(x)}{h}=\frac{\frac{3(x+h)}{2-(x+h)}-\frac{3x}{2-x}}{h}=\frac{\frac{3x+3h}{2-x-h}-\frac{3x}{2-x}}{h}=\frac{\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{(2-x-h)(2-x)}}{h}\]
\[=\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{h(2-x-h)(2-x)}=\frac{6x-3x^2+6h-3hx-6x+3x^2+3xh}{h(4-2x-2x+x^2-2h+xh)}\]
\[=\frac{6h}{h(4-2x-2x+x^2-2h+xh)}=\frac{6}{4-4x+x^2-2h+xh}=\frac{6}{x^2-4x+xh-2h+4}\]

Answer 10

um i cant see the whole thing

Answer 11

\[=\frac{6h}{h(4-2x-2x+x^2-2h+xh)}=\frac{6}{4-4x+x^2-2h+xh}\]
\[=\frac{6}{x^2-4x+xh-2h+4}\]
This looks like derivative work to find the tangent to \(f(x)\) at any x value so ill simple it up for you:
\[\lim_{h\rightarrow0}{\frac{6}{x^2-4x+xh-2h+4}}=\frac{6}{x^2-4x+4}=\frac{6}{(x-2)^2}\]

Answer 12

Oh damn. Lol let me rewrite it much quicker:

Answer 13

i dont need the derivative, just simplified

Answer 14

how did u just get a 6 h on top?

Answer 15

I cancelled out all the other ones so like \(-6x^2\) and \(6x^2\) cancel eachother out

Answer 16

let me see if i get that

Answer 17

\[f(x)=\frac{3x}{2-x}\]
\[\eqalign{
\frac{f(x+h)-f(x)}{h}&=\frac{\frac{3(x+h)}{2-(x+h)}-\frac{3x}{2-x}}{h} \\
&=\frac{\frac{3x+3h}{2-x-h}-\frac{3x}{2-x}}{h} \\
&=\frac{\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{(2-x-h)(2-x)}}{h} \\
&=\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{h(2-x-h)(2-x)} \\
&=\frac{6x-3x^2+6h-3hx-6x+3x^2+3xh}{h(4-2x-2x+x^2-2h+xh)} \\
&=\frac{6h}{h(4-2x-2x+x^2-2h+xh)} \\
&=\frac{6}{4-4x+x^2-2h+xh} \\
&=\frac{6}{x^2-4x+xh-2h+4} \\}\]

Answer 18

hmm

Answer 19

Did you get it?