# Thread: Finding Domain and Range

1. ## Finding Domain and Range

5y(x-109x-3) = 2(5x+3)
I already got the domain which is x≠ 1,3.

My problem is on how to get the range. This is what I started.
5y(x-109x-3) = 2(5x+3)
5y(x^2-4x+3) = 10x+15
I stuck until here. I don't know if I did the right move.
A friend told me that the range is y≤-5 but he could not show me the whole solution.

2. Originally Posted by vldo
5y(x-109x-3) = 2(5x+3)
I already got the domain which is x≠ 1,3.

My problem is on how to get the range. This is what I started.
5y(x-109x-3) = 2(5x+3)
5y(x^2-4x+3) = 10x+15
I stuck until here. I don't know if I did the right move.
A friend told me that the range is y≤-5 but he could not show me the whole solution.
First off, I assume you are talking about the last equation you've written and not the first two. The domain would be wrong if you were talking about the first two.

In calculus, you can find the domain by analyzing the derivatives of your function.

First, find $\frac{dy}{dx}$ and set it equal to zero to find all local maxima and minima. A first derivative test will tell you the function is decreasing from x=negative infinity to x=-3 (where it hits a minimum at (-3,-0.2)), then increasing untiil x=1 (there is a discontinuity at x=1), then increases until x=1.8 (where there is a maximum at (1.8,-5)), then decreases until x=3 (there is a disconuity at x=3), then decreases over the remainder of its domain.

Can you picture this in your mind? Or even draw a quick sketch of this behavior?

The only real question is whether the functional values pass between -0.2 and -5 when it is decreasing as x gets larger. Looking at the original function there is a horizontal asymptote at y=0, so it can't pass through these values.

You can go through the same analysis of the second derivative to determine the intervals where the function is concave up and down. But the first derivative test will really tell you what you need to know for this function.

The range is -0.2<=y<=-5.

3. ## domain & range

Originally Posted by vldo
5y(x-109x-3) = 2(5x+3)
I already got the domain which is x≠ 1,3.

My problem is on how to get the range. This is what I started.
5y(x-109x-3) = 2(5x+3)
5y(x^2-4x+3) = 10x+15
I stuck until here. I don't know if I did the right move.
A friend told me that the range is y≤-5 but he could not show me the whole solution.

After looking at my keyboard - to see what likely typos you made - I assume you typed 0 for ) and 9 for (. So you must mean:
What is the range of 5y(x-1)(x-3) = 2(5x+3) ?

Now solve for y - no need to expand (x-1)(x-3).

$\displaystyle y=\left({2\over 5}\right){{5x+3}\over{(x-1)(x-3)}}$

You have the domain right.

There is a horizontal asymptote: y=0. However, this doesn't mean $y\ne0$. It only gives the behavior of y when $|x|\to +\infty$. In fact y=0 when $x = -{3\over5}$.

As mentioned elsewhere, there is a local minimum at (-3,-0.2), and a local maximum at (1.8, -5). y does not take on any values between -5 and -0.2 .

Therefore, y<=-5 OR y>=-0.2. In interval notation this is:

Range = $(-\infty,\ -5] \cup [-0.2,+\infty)$
.