2x/(4x^2-2x)

Question

2x/(4x^2-2x)

jhannybean · Answer

what do you get when you factor out a 2x in the denominator?

anonymous · Answer

attachment

jhannybean · Answer

no...

anonymous · Answer

1

jhannybean · Answer

$$\large \frac{2x}{4x^2-2x}$$ when you pull out a 2x at the bottom, which is a common multiple, what does it factor out to?

anonymous · Answer

I have to simplify and state restrictions and do not know how to do this

anonymous · Answer

are you going to reply?

jhannybean · Answer

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 · Answer

yes

jhannybean · Answer

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 · Answer

2X

jhannybean · Answer

good job. now we can cancel them out as so: $$\large \frac{\cancel{2x}}{\cancel{2x}(2x-1)} = \frac{1}{2x-1}$$

jhannybean · Answer

you see how we still have an x in the denominator? We can evaluate that by equaling it to 0.

anonymous · Answer

ok

jhannybean · Answer

$$\large 2x-1 = 0$$ can you solve for x?

anonymous · Answer

x=1/2

anonymous · Answer

is that the answer

jhannybean · Answer

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 · Answer

Can you tell me what you would get if you plugged x=1/2 into 

1/ 2x-1 ?

anonymous · Answer

attachment

jhannybean · Answer

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 · Answer

See how that works? because we have 1-1 in the denominator.

jhannybean · Answer

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 · Answer

So our restriction is $$\large (-\infty, 1/2) \cup (1/2 , \infty)$$

jhannybean · Answer

Do you understand what i'm saying?

anonymous · Answer

no

anonymous · Answer

we do not do it like this in class