anonymous
  • anonymous
2x/(4x^2-2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
what do you get when you factor out a 2x in the denominator?
anonymous
  • anonymous
0
Jhannybean
  • Jhannybean
no...

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anonymous
  • anonymous
1
Jhannybean
  • Jhannybean
\[\large \frac{2x}{4x^2-2x}\] when you pull out a 2x at the bottom, which is a common multiple, what does it factor out to?
anonymous
  • anonymous
I have to simplify and state restrictions and do not know how to do this
anonymous
  • anonymous
are you going to reply?
Jhannybean
  • Jhannybean
ok. Lets see. \[\large \frac{2x}{2x(2x-1)}\] do you see that if we multiply the 2x with the (2x - 1) inside the parenthesis that we will end up with 4x^2-2x?
anonymous
  • anonymous
yes
Jhannybean
  • Jhannybean
Okay, now that we have factored out a 2x (this si what i meant by factoring out a common multiple) you can cancel like terms. There are two like terms I see here, one in the denominator, and one in the numerator, can you tell me what they are?
anonymous
  • anonymous
2X
Jhannybean
  • Jhannybean
good job. now we can cancel them out as so: \[\large \frac{\cancel{2x}}{\cancel{2x}(2x-1)} = \frac{1}{2x-1}\]
Jhannybean
  • Jhannybean
you see how we still have an x in the denominator? We can evaluate that by equaling it to 0.
anonymous
  • anonymous
ok
Jhannybean
  • Jhannybean
\[\large 2x-1 = 0\] can you solve for x?
anonymous
  • anonymous
x=1/2
anonymous
  • anonymous
is that the answer
Jhannybean
  • Jhannybean
good job. now we've clarified that if x= 1/2 in our denominator . we know if we plug it back into our simplified equation, we would get a function that didn't exist, because we would have a 0 in our denominator.
Jhannybean
  • Jhannybean
Can you tell me what you would get if you plugged x=1/2 into 1/ 2x-1 ?
anonymous
  • anonymous
0
Jhannybean
  • Jhannybean
no, you would only get 0 if it was 2x-1/1 \[\large \frac{1}{2(\frac{1}{2})- 1} = \frac{1}{0} = \text {undefined}\]
Jhannybean
  • Jhannybean
See how that works? because we have 1-1 in the denominator.
Jhannybean
  • Jhannybean
so when a function = 0 OR is undefined, that means we have a restriction in our function. the function will be continuous from negative infinity ALL THE WAY up to 1/2, where there will either be a split (asymptote) a break, or a sharp turn in the graph, and the function will then continue past x=1/2 to positive infinity.
Jhannybean
  • Jhannybean
So our restriction is \[\large (-\infty, 1/2) \cup (1/2 , \infty)\]
Jhannybean
  • Jhannybean
Do you understand what i'm saying?
anonymous
  • anonymous
no
anonymous
  • anonymous
we do not do it like this in class

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