sketch 3x-2/sqrt[x^3+8], and find domain, range

[JuventudEnExtasis]

I have been trying all day and I haven't been able to figure this homework problem out. Math is my weakest area. Can someone help me figure it out and maybe tell me how to do it? I would really appreciate it

1. Sketch a graph of the function: f(x) = 3x-2/√x^3+8

Find the function's domain and range.

 

[galactus]

Don't you have a graphing calculator?.

As for the domain and range.

Look at your radical. You can't divide by 0 or have negatives insde the radical. What results in 0 and negatives in the radical?. Your domain is everything but that.

Think about x=-2 maybe.

 

[JuventudEnExtasis]

i'm just having trouble finding the range...i figured out how to graph it and i got the domain down but the range is hard

 

[galactus]

The range isn't hard. Think of it this way. The domain is what you put in

and the range is what comes out. You know you can't use \(\displaystyle x\leq{-2}\) because that'll result in 0 or a negative in the radical.

If you put in something that is very close to -2, but not exactly -2, you get larger and larger negative values. This would suggest the range is \(\displaystyle {-\infty}\), from that direction. Now try values geting farther from -2. As x gets larger the function approaches 0. See?. Graph your problem. You'll see.

Can you figure out what the range is now?.

 

[soroban]

Hello, JuventudEnExtasis!

I saw your post at another site and no one there has solved it either.

1. Sketch a graph of the function: \(\displaystyle \L f(x) \:= \:\frac{3x\,-\,2}{\sqrt{x^3\,+\,8}}\)

Find the its domain and range.

You already found that the domain is: \(\displaystyle \-2,\,\infty)\)

It has x-intercept \(\displaystyle \left(\frac{2}{3},\,0\right)\) and y-intercept \(\displaystyle (0,\,-1)\)

There is a vertical asymptote \(\displaystyle x\,=\,-2\) where the curve approaches -\(\displaystyle \infty\).


As \(\displaystyle x\to\infty\), the function approaches: \(\displaystyle \L\,\frac{3x}{\sqrt{x^3}} \:=\:\frac{3}{\sqrt{x}}\) which approaches \(\displaystyle 0\).
. . Hence, \(\displaystyle y\,=\,0\) (x-axis) is a horizontal asymptote.

So, the graph looks like this:

           :    |
           :    |              *
           :    |        *           *
           :    |    *                        * 
        ---:----+-*------------------------------
           :    *
           :  * |
           : *  |
           :    |
           :*   |
           :    |

The function has a maximum point near \(\displaystyle x\,=\,3.4\)
. . but locating its exact value is difficult.

Good luck!