# Range homework

1. Sep 4, 2008

### BoundByAxioms

1. The problem statement, all variables and given/known data
$$\frac{x-5}{x^{2}-9}$$

2. Relevant equations
Usual range-finding methods. For this one, I thought that if the exponent of the numerator is smaller than that of the denominator, then there should be an asymptote at y=0, aka the range doesn't include 0. I think there is a special case for when the exponent on the numerator is exactly one greater than that of the denominator, but I'm not completely sure...

3. The attempt at a solution
Already explained my reasoning. But, range should be y$$\neq$$0.

2. Sep 4, 2008

### Defennder

Re: Range

Firstly, assume the largest possible range, R (any real number). Let R = (your expression). Bring the denominator over to R and you'll get a quadratic equation. Remember that you are assuming that the domain is any real number except +/-3. With that in mind, figure out what the quadratic expression must satisfy in order for the domain to be as assumed.

3. Sep 4, 2008

### BoundByAxioms

Re: Range

Ok let's see:

R=$$\frac{x-5}{x^{2}-9}$$. Then, R(x^{2}-9)=x-5. Then
Rx^{2}-x-9R+5=0. At this point I'm not sure what I'm doing, so I'll give
it a guess: Since x$$\neq$$+/-3, I'll plug (-3) and (3) into x and
see what I get. (3R)^{2}-3-9R+5=0 which simplifies to 9R^{2}-9R=-2.
R^{2}-R+$$\frac{2}{9}$$=0. Using the quadradic formula:
$$\frac{1 (+/-)\sqrt{(-1)^{2}-4(1)(\frac{-2}{9}} }{2}$$. So it
looks like x=$$\frac{1}{2}$$ +/- $$\frac{1}{3}$$, so
x=$$\frac{5}{6}$$ and x=$$\frac{1}{6}$$. So the range can't
include those two values? By the graph it looks like the range doesn't
include anything between those two values. And why isn't there an
asymptote at y=0?

4. Sep 24, 2008

### guynoone

Re: Range

Bumping thread since I need to know the answer also.

5. Sep 24, 2008

### Mentallic

Re: Range

There is an asymptote at the x-axis, but as $$x\rightarrow\infty$$ : $$y\rightarrow0^{+}$$.

The function crosses from the negative values into the positive (thus intersecting the x-axis at x=3) but then treats y=0 as an asymptote from there on.

I am also curious for the answer here.

6. Sep 24, 2008

### HallsofIvy

Staff Emeritus
Re: Range

The range certainly does include 0. It should be obvious that y(5)= (5-5)/(25-9)= 0.

y= (x-5)/[(x-3)(x+ 3)]. There are vertical asymptotes at x= 3 and -3. For x< -3, all three factors are negative so y is always negative. The graph approaches y= 0 from below as x goes to -infinity, then goes down to -infinity as x approaches 3 from below. y is very large for x slightly larger than -3, reaches a local minimum at about x= -1, then goes to infinity as x goes to 3 from below. For x slightly greater than 3, y is a large negative number, crosses the x axis at x= 5, reaches a local maximum at about x= 11 then decreases toward 0 as x goes to infinity.

The range is all real numbers, including 0.

7. Sep 24, 2008

### Defennder

Re: Range

Hi Halls,

I'm not seeing why the range is all real numbers. I tried out what I suggested earlier and I determined the range of values to be either <0.0574 or >0.387. Which means values like 0.1 isn't found in the range. If we consider x to be restricted to the reals (I assume it is), then there isn't any real number x such that the expression = 0.1.

8. Sep 24, 2008

### BoundByAxioms

Re: Range

Take a look at the graph and you'll see that not all real numbers are within its range.

What I've been told is that there is always an asymptote at y=0 whenever you have something like f(x)=$$\frac{x^{a}}{x^{b}}$$ where a<b. When is that true? Because clearly it's not true for this particular function.

9. Sep 25, 2008

### HallsofIvy

Staff Emeritus
Re: Range

Yes, you are right! I didn't look at it closely enough before- there a small horizontal lane the graph does not go into- very difficult to get any specific values but differentiating and finding the local max and min should do it.

That certainly is true of this function. What you were probably NOT told is that a graph can't cross a horizontal asymptotote.

Important point: "there should be an asymptote at y=0, aka the range doesn't include 0" is WRONG. Saying y= a is a horizontal asymptote only tells you that the graph approaches y= a as x goes to infinity or negative infinity. It does NOT tell you anything about what happens for other values of x. In particular it does not tell you that the graph does not cross y= a.

Graph y= x/(x2+ 1). That's a continuous function that has y= 0 as a horizontal asymptote but is 0 for x= 0.

Last edited: Sep 25, 2008
10. Sep 26, 2008

Re: Range

If $$t$$ is a real number in the range of this function, then

$$t = \frac{x-5}{x^2-9}$$

To find the set of numbers that can be used for $$t$$

\begin{align*} t (x^2 - 9) & = x-5 \\ tx^2 - 9t - x + 5 & = 0\\ tx^2 - x + (5-9t) & = 0 \end{align*}

If the determinant of the final quadratic is negative there are no solutions for $$x$$. Since the determinant will depend on $$t$$, we can find the values that are not in the range.

\begin{align*} D(t) & = (-1)^2 - 4(t)(5-9t) \\ & = 1 - 20t + 36t^2 \end{align*}

The zeros of the quadratic in $$t$$ are $${36}/{72} = 1/2$$ and $$4/{72} = 1/{18}$$. Since the graph of $$D(t)$$ is a parabola that opens upward, the vertex is the minimum point.

\begin{align*} t & = \frac{-(-20)}{72} = \frac{20}{72} = \frac 5{18}\\ D(5/{18}) & \approx -1.8 \end{align*}

Since $$D(t)$$ is negative at the vertex, it is negative throughout the interval

$$\left(\frac{4}{72}, \frac 1 2 \right)$$

The range consists of all real numbers except for those in this interval

11. Sep 27, 2008

### HallsofIvy

Staff Emeritus
Re: Range

Well done, statdad, though not necessary to post it twice!. Oh, and the word you want is "discriminant", not "determinant".

12. Sep 27, 2008