Find the domain restrictions

Question

btaylor · Answer

of...?

anonymous · Answer

hold on one sec let me post it $$2v(2v^3 + 3) / 5v-4$$

btaylor · Answer

Alright. So a domain restriction is some value $v$ where the function is not defined.

btaylor · Answer

These occur when a denominator equals 0 (usually a hole or an asymptote), or when you have the square root of a number less than zero. Which case applies here?

anonymous · Answer

the denominator equals zero?

btaylor · Answer

yes. So, in your problem, $5v-4\neq 0$. What $v$ makes this true?

anonymous · Answer

attachment

btaylor · Answer

Sorry, I should've phrased that better. What $v$ would make the denominator equal to 0?

anonymous · Answer

5v-4 cause they would cancle each other out

btaylor · Answer

Here's how I would start it. I'm using $\neq$ instead of = because I want to find what $v$ cannot equal.

btaylor · Answer

$5v-4 \neq 0 \rightarrow 5v \neq 4$\ What can't $v$ be?

anonymous · Answer

dont you want the opposite of it

btaylor · Answer

no, you want that. Because when v =4/5, then the denominator equals , so the function is not defined.

btaylor · Answer

So your domain restriction will be $v \neq 4/5$

anonymous · Answer

how do i get it to that?

btaylor · Answer

From $5v-4 \neq 0$, add 4 to both sides. You get $5v \neq 4$. Then divide both sides by 5 to get $v \neq 4/5$.

anonymous · Answer

thanks so much

btaylor · Answer

no problem