1. ## Continuity Problems

I'm having a heck of a time figuring out how they got an answer to a problem. Maybe you guys can help me out?

Find values of X, if any at which f is not continuous.

X/2x^2+x books answer is 0 and -1/2

Wouldn't I factor this denominator? I understand why its 0 and -1/2 has it makes the bottom denominator equal to 0 but how did they get to that answer? I know just plugging in 0 will prove to make the denominator 0 but how did they do the -1/2? Wouldn't they factor? Thanks

Another one i'm having problems with is

radical x-8 with index 3. Don't even know how to start this one.

THanks

2. Originally Posted by dan87951
I'm having a heck of a time figuring out how they got an answer to a problem. Maybe you guys can help me out?

Find values of X, if any at which f is not continuous.

X/2x^2+x books answer is 0 and -1/2

Wouldn't I factor this denominator? I understand why its 0 and -1/2 has it makes the bottom denominator equal to 0 but how did they get to that answer? I know just plugging in 0 will prove to make the denominator 0 but how did they do the -1/2? Wouldn't they factor? Thanks

Another one i'm having problems with is

radical x-8 with index 3. Don't even know how to start this one.

THanks
If you mean $\displaystyle \frac{x}{2x^2 + x}$

Then, you have a discontinuity whenever the entire denominator $\displaystyle = 0.$

That is, when $\displaystyle 2x^2 + x = 0$

$\displaystyle \implies x(2x + 1) = 0$

$\displaystyle \underbrace{x}\underbrace{(2x + 1)} = 0$

Which $\displaystyle =0$ when either under-bracketed part $\displaystyle =0.$

i.e., $\displaystyle x= 0$ or, when $\displaystyle 2x+1 = 0 \implies x=-\frac{1}{2}$

3. Originally Posted by dan87951
I'm having a heck of a time figuring out how they got an answer to a problem. Maybe you guys can help me out?

Find values of X, if any at which f is not continuous.

X/2x^2+x books answer is 0 and -1/2

Wouldn't I factor this denominator? I understand why its 0 and -1/2 has it makes the bottom denominator equal to 0 but how did they get to that answer? I know just plugging in 0 will prove to make the denominator 0 but how did they do the -1/2? Wouldn't they factor? Thanks

Another one i'm having problems with is

radical x-8 with index 3. Don't even know how to start this one.

THanks
Dear Anonymous1,

$\displaystyle \frac{x}{2x^2-x}=\frac{x}{x(2x-1)}=\frac{x}{2x\left(x-\frac{1}{2}\right)}$

4. I think its just easier for me to use the quadratic formula for this this type of problem. I don't understand the algebra you guys are doing. Is there a website that will help me understand this?

5. Originally Posted by dan87951
I think its just easier for me to use the quadratic formula for this this type of problem. I don't understand the algebra you guys are doing. Is there a website that will help me understand this?
Dear dan87951,

What do you don't understand? I have factorized the denominator.

$\displaystyle 2x^2-x=x(2x-1)$; taking x outside the bracket

$\displaystyle x(2x-1)=2x\left(x-\frac{1}{2}\right)$; taking 2 outside the bracket