Hiroromon
  • Hiroromon
if f(x) = (3x)/(-x +2) find (f(x+h) - f(x))/h and simplify completely
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[f(x)=\frac{3x}{2-x}~~\Rightarrow~~f(x+h)=\frac{3(x+h)}{2-(x+h)}\] So you have \[\frac{f(x+h)-f(x)}{h}=\frac{1}{h}\left(\frac{3(x+h)}{2-(x+h)}-\frac{3x}{2-x}\right)\]
Hiroromon
  • Hiroromon
do i just distribute everything out now?
anonymous
  • anonymous
Finding a common denominator first might help. You should combine the fractions.

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Hiroromon
  • Hiroromon
um ok
anonymous
  • anonymous
Looks like @KeithAfasCalcLover is in the process of typing it all out. I'll leave it to him.
Hiroromon
  • Hiroromon
|dw:1378522164610:dw|
Hiroromon
  • Hiroromon
can i cross out the (-2x+2h)?
Hiroromon
  • Hiroromon
and (2-x)?
Hiroromon
  • Hiroromon
yes no?
anonymous
  • anonymous
\[f(x)=\frac{3x}{2-x}\] \[\frac{f(x+h)-f(x)}{h}=\frac{\frac{3(x+h)}{2-(x+h)}-\frac{3x}{2-x}}{h}=\frac{\frac{3x+3h}{2-x-h}-\frac{3x}{2-x}}{h}=\frac{\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{(2-x-h)(2-x)}}{h}\] \[=\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{h(2-x-h)(2-x)}=\frac{6x-3x^2+6h-3hx-6x+3x^2+3xh}{h(4-2x-2x+x^2-2h+xh)}\] \[=\frac{6h}{h(4-2x-2x+x^2-2h+xh)}=\frac{6}{4-4x+x^2-2h+xh}=\frac{6}{x^2-4x+xh-2h+4}\]
Hiroromon
  • Hiroromon
um i cant see the whole thing
anonymous
  • anonymous
\[=\frac{6h}{h(4-2x-2x+x^2-2h+xh)}=\frac{6}{4-4x+x^2-2h+xh}\] \[=\frac{6}{x^2-4x+xh-2h+4}\] This looks like derivative work to find the tangent to \(f(x)\) at any x value so ill simple it up for you: \[\lim_{h\rightarrow0}{\frac{6}{x^2-4x+xh-2h+4}}=\frac{6}{x^2-4x+4}=\frac{6}{(x-2)^2}\]
anonymous
  • anonymous
Oh damn. Lol let me rewrite it much quicker:
Hiroromon
  • Hiroromon
i dont need the derivative, just simplified
Hiroromon
  • Hiroromon
how did u just get a 6 h on top?
anonymous
  • anonymous
I cancelled out all the other ones so like \(-6x^2\) and \(6x^2\) cancel eachother out
Hiroromon
  • Hiroromon
let me see if i get that
anonymous
  • anonymous
\[f(x)=\frac{3x}{2-x}\] \[\eqalign{ \frac{f(x+h)-f(x)}{h}&=\frac{\frac{3(x+h)}{2-(x+h)}-\frac{3x}{2-x}}{h} \\ &=\frac{\frac{3x+3h}{2-x-h}-\frac{3x}{2-x}}{h} \\ &=\frac{\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{(2-x-h)(2-x)}}{h} \\ &=\frac{(3x+3h)(2-x)-(3x)(2-x-h)}{h(2-x-h)(2-x)} \\ &=\frac{6x-3x^2+6h-3hx-6x+3x^2+3xh}{h(4-2x-2x+x^2-2h+xh)} \\ &=\frac{6h}{h(4-2x-2x+x^2-2h+xh)} \\ &=\frac{6}{4-4x+x^2-2h+xh} \\ &=\frac{6}{x^2-4x+xh-2h+4} \\}\]
Hiroromon
  • Hiroromon
hmm
anonymous
  • anonymous
Did you get it?

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