Continuity Problems

May 2010
5
0
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
 
Nov 2009
517
130
Big Red, NY
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 \(\displaystyle 2x+1 = 0 \implies x=-\frac{1}{2}\)
 
Dec 2009
872
381
1111
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,

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

Hope this will help you.
 
May 2010
5
0
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?
 
Dec 2009
872
381
1111
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

Hope this will help you.