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 \(\displaystyle 2x+1 = 0 \implies x=-\frac{1}{2}\)