# Thread: help with the domain of the function

1. ## help with the domain of the function

can someone please explain to find the domain of this function.

I keep getting:

x^2(-2x+21)(-67x+60)
x^2(21/2)(60/67)

which i type in my homework & is wrong.
does anyone know what to do?

2. Originally Posted by lsnyder
can someone please explain to find the domain of this function.

I keep getting:

x^2(-2x+21)(-67x+60)
x^2(21/2)(60/67)

which i type in my homework & is wrong.
does anyone know what to do?
The only limit imposed on this function is that the denominator CANNOT be zero.

So I recommend you factorise the denominator and find out which values of x lead the denominator to be 0. The domain of the function is all values of x, apart from the 3 values of x that lead the denominator to be 0.

3. Isnt the domain of a function where you find what numbers you can use to that don't make the problem undefined?

I think you're supposed to start out by factoring the top. You can take x out and you have
x(2x^2-1x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(x-3/2)(x-4)(x-5)

(-infinity,4) u (4,5) u (5,infinity)

Sorry, I dont know latex yet.

4. Originally Posted by ninobrn99
Isnt the domain of a function where you find what numbers you can use to that don't make the problem undefined?

I think you're supposed to start out by factoring the top. You can take x out and you have
x(2x^2-1x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(-2x^3+21x^2-67x+60)
=x(x+1)(2x-3)/(x-3/2)(x-4)(x-5)

(-infinity,4) u (4,5) u (5,infinity)

Sorry, I dont know latex yet.
You don't have to factor the top. The top is irrelevant to the problem.

You just have to factorise the bottom and that gives you your limits. As ytou rightly said.

$\text{denominator} = \bigg(x - \frac{3}{2}\bigg)\bigg(x -4\bigg)\bigg(x-5\bigg)$

Hence $x \in \bigg(- \infty, \frac{3}{2} \bigg) \cup \bigg( \frac{3}{2}, 4 \bigg) \cup \bigg(4, 5 \bigg) \cup \bigg(5, \infty \bigg)$

5. Originally Posted by Mush
You don't have to factor the top. The top is irrelevant to the problem.

You just have to factorise the bottom and that gives you your limits. As ytou rightly said.

$\text{denominator} = \bigg(x - \frac{3}{2}\bigg)\bigg(x -4\bigg)\bigg(x-5\bigg)$

Hence $x \in \bigg(- \infty, \frac{3}{2} \bigg) \cup \bigg( \frac{3}{2}, 4 \bigg) \cup \bigg(4, 5 \bigg) \cup \bigg(5, \infty \bigg)$
yeah, sorry. Im just so used to factoring everything. Just creating more work for myself, but yea it is only the bottom.

6. can some one explain how to find the hole and vertical acytope please.

The root(s) of , in increasing order, is/are: , , .

(last one does not exist, I know so far.)

has one hole at the point: ( , ).

7. Originally Posted by lsnyder
can some one explain how to find the hole and vertical acytope please.

The root(s) of , in increasing order, is/are: , , .

(last one does not exist, I know so far.)

has one hole at the point: ( , ).
What do you mean by hole?

A vertical asymptote is when the curve trails off to plus or minus infinity at a certain point, because the function is not defined at that point.

Your function is not defined for $x = \frac{3}{2}, x = 4, x = 5$. So there will by 3 vertical asymptotes. Whether the curve tends towards these positively or negatively depends on a table of signs.

I have no idea what a 'hole' is though.

8. Just discovered what a hole is lol.

A hole is when the zero of the numerator is equal to the zero on the denominator.

So for this one we DO need to factorise both top and bottom, and that gives us:

$\frac{x(x+1)(2x-3)}{\big(x-\frac{3}{2}\big)(x-4)(x-5)}$

Which can be written:

$\frac{x(x+1)(2x-3)}{(2x-3)(x-4)(x-5)}$

And you can see that the function has a zero on the top and bottom when $x = \frac{3}{2}$. So that is the x co-ordinate of your hole.

But to be honest, this is a trivial matter. We already knew there would be a 'gap' in the function that point due to the fact that the VA is there. I see no use in this 'hole' business.

9. okay this might of been answered, but how do i find the other hole?

because once again, i don't get it