removable discontinuity help?

Question

jessicawade · Answer

for the function $$f(x)= 5x^3-x \over 2x^3$$ identify any removable discontinuity

jessicawade · Answer

There are removable discontinuities at x = –1 and x = 1.  
	There is a removable discontinuity at x = 0.
	There is a removable discontinuity at x= 5/2
	There are no removable discontinuities for this function.

jessicawade · Answer

@dtan5457

anonymous · Answer

Okay, in which cases you would have a discontinuity?
When your function goes "indetermined", in fractions the most common indetermination es a/0 (divide by zero), so...what do you think?

jhannybean · Answer

Think of a removable discontinuity as removing a bad friend from your life. they still exist, but theyre not around you. HAH

jhannybean · Answer

$$f(x) = \frac{5x^3-x}{2x^3}$$

jhannybean · Answer

What is the LCM in the numerator, between $x^3$ and \(x)?

jessicawade · Answer

yeah thats it lol and i dont know how to find it lol

jhannybean · Answer

$x$*?

jessicawade · Answer

hmmm

jhannybean · Answer

which one is smaller between the two.

jessicawade · Answer

2 right?

jhannybean · Answer

No, you have $\large x^{\color{red}{3}}$ and $\large x^{\color{red}{1}}$

jhannybean · Answer

Which one is smaller between these two variables,

jessicawade · Answer

x

jhannybean · Answer

Good. so if you factor out an x, what do you end up with in the numerator?

jessicawade · Answer

5x^3

jessicawade · Answer

what next?

jhannybean · Answer

Not quite. $x(5x^2-1)$ If you multiplied the x back in, you would end up with $5x^3 -x$

jhannybean · Answer

So you have $f(x) = \dfrac{x(5x^2 -1)}{2x^3}$

jessicawade · Answer

oh. lol

jessicawade · Answer

then what do i do?

jhannybean · Answer

Cancel out like terms. you have $x^1$ in the numerator and $x^3$ in the denominator. this will reduce to $\dfrac{x^1}{x^3} = x^{1-3}= x^{-2} = \dfrac{1}{x^2}$

jessicawade · Answer

ok

jhannybean · Answer

What will the expression simplify to?

jessicawade · Answer

hmm

jessicawade · Answer

not sure i wanna say something dumb like 1 lol

jessicawade · Answer

@Jhannybean

jhannybean · Answer

Well, you canceled out the x term in the numerator and were left with $x^2$ in the denominator, so what is your function going to look like then?

jessicawade · Answer

just x^2 if the numerator is gone right? sorry this is old work for me

jhannybean · Answer

We still have the $(5x^2 -1)$ left though..

jessicawade · Answer

oh. o.o

jessicawade · Answer

hmm im not sure

michele_laino · Answer

I think as @Jhannybean  said your function can be rewrite as below:
$$f(x)=\frac{ 5x ^{2} -1}{ 2 x ^{2} }$$
so your discontinuity is not removable

michele_laino · Answer

please note that you have a discontnuity at x=0

jessicawade · Answer

so it does have one at x=0?

jessicawade · Answer

wait. 
There are removable discontinuities at x = –1 and x = 1.
 There is a removable discontinuity at x = 0.
 There is a removable discontinuity at x= 5/2 
There are no removable discontinuities for this function.

jessicawade · Answer

is it at x=0 or does it not have any removable discon.

jessicawade · Answer

@Michele_Laino

michele_laino · Answer

the last option, namely there are no  removable discontinuities

jessicawade · Answer

oh ok