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So i understand the whole concept of getting the same denominator using the rules. But the problem i have is plugging these values into the formula itself. The variable with the top numerator is giving me trouble. I have no problem with plugging in any other functions whatsoever, besides these fractions. If someone could show me how they plugged in the values then i can do the rest.

Veteran  Apr 14, 2017
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2+0 Answers

 #1
avatar+4174 
+3

I think I can do this one.

\(k(\textcolor{lime}{x})=\frac{10\textcolor{lime}{x}}{\textcolor{lime}{x}-2}\)

 

So..

 

\(k(\textcolor{lime}{x+h}) = \frac{10(\textcolor{lime}{x+h})}{(\textcolor{lime}{x+h})-2} \\~\\ k(\textcolor{lime}{x}) = \frac{10\textcolor{lime}{x}}{\textcolor{lime}{x}-2} \\~\\ h = h\)

 

Now just sub these in.

 

\(\large{\frac{k(x+h)-k(x)}{h}=\frac{\frac{10(x+h)}{(x+h)-2}-\frac{10x}{x-2}}{h}}\)

 

Then simplify, which seems like it might be kind of hard!

 

 

*edit* This is what I get for the simplifying part:

 

\(=(\frac{1}{h})(\frac{10(x+h)}{x+h-2}-\frac{10x}{x-2}) \\~\\ =(\frac{1}{h})(\frac{10(x+h)\cdot(x-2)}{(x+h-2)\cdot(x-2)}-\frac{10x\cdot(x+h-2)}{(x-2)\cdot(x+h-2)}) \\~\\ =(\frac{1}{h})(\frac{10(x+h)(x-2)-10x(x+h-2)}{(x+h-2)(x-2)}) \\~\\ =(\frac{1}{h})(\frac{(10x^2-20x+10hx-20h)-(10x^2+10xh-20x)}{x^2-2x+hx-2h-2x+4}) \\~\\ =(\frac{1}{h})(\frac{10x^2-20x+10hx-20h-10x^2-10xh+20x}{x^2-4x+hx-2h+4}) \\~\\ =(\frac{1}{h})(\frac{-20h}{x^2-4x+hx-2h+4}) \\~\\ =-\frac{20}{x^2-4x+hx-2h+4}\)

 

Just in case that helps any! :)

hectictar  Apr 14, 2017
edited by hectictar  Apr 14, 2017
 #2
avatar+223 
+1

awsome, this helped me find the problem i had from the start of it.

Veteran  Apr 14, 2017

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