Question

g(x) = sqrt(x+5) / x^2+x=30

Answer 1

need to find the domain in interval notation please step by step would like to see how this is done.

Answer 2

I assume that the denominator is supposed to be x^2+x+30?

Answer 3

x^2+x-30 , my mistake.

Answer 4

First note that the domain of \(\sqrt{x}\) is \([0,\infty)\). So what is the domain of \(\sqrt{x+5}\)? Secondly, note that \(x^2+x-30 = (x+6)(x-5)\) and hence \((x+6)(x-5)=0\) when \(x=-6,5\). Thus, the domain of \(\dfrac{1}{x^2+x-30}\) is \((-\infty,-6)\cup(-6,5)\cup(5,\infty)\). Hence the domain of \(\dfrac{\sqrt{x+5}}{x^2+x-30}\) is the interval of overlap between the individual pieces.

Can you take things from here? :-)

Answer 5

By individual pieces, I mean the domains of \(\sqrt{x+5}\) and \(\dfrac{1}{x^2+x-30}\).

Answer 6

i dont understand the x=-6,5

Answer 7

im kind of confused of where the 6 and 5 came from

Answer 8

When you factor x^2+x-30 you get (x+6)(x-5), that's where he got it from.

Answer 9

And since it's in the denominator, you want to be wary of the values that cause you to divide by zero.

Answer 10

i dont understand the very last part what does that mean its the overlap?

Answer 11

What I mean by that is the following. The \(\sqrt{x+5}\) function has the domain \([-5,\infty)\) and the function \(\dfrac{1}{x^2+x-30}\) has the domain \((-\infty,-6)\cup(-6,5)\cup(5,\infty)\). However, when you put multiply them together to get \(\dfrac{\sqrt{x+5}}{x^2+x-30}\), the resulting domain is the intersection of the two domains we found earlier (i.e. where the two domains overlap). So what is the resulting domain?

I hope this clarifies things! :-)

Answer 12

what does the intersection mean? i dont understand where the 2 overlap

Answer 13

Consider the domains on the following number lines below:



If you drop down a line starting where the two domains of the first two functions start to overlap, you see that the function of interest \(\dfrac{\sqrt{x+5}}{x^2+x-30}\) must be defined for \(x\geq -5\) since \(\sqrt{x+5}\) isn't defined for negative values (i.e. when \(x<-5\)). The next thing to note that \(x\neq 5\) since the denominator of \(\dfrac{\sqrt{x+5}}{x^2+x-30}\) is zero there. Other than that, the function is defined for all other \(x\geq -5\). Thus, the domain of \(\dfrac{\sqrt{x+5}}{x^2+x-30}\) is \([-5,5)\cup(5,\infty)\).

I hope this makes sense! :-)