Is it possible to find the range of this function?

[vcsharp2003]

Homework Statement

Find the domain and range of following function: $f(x) =\frac {x^2 -1} {x-5}$.

Relevant Equations

None

I can get the domain, but getting the range seems impossible.

Domain
$$x-5=0$$
$$x =5$$
$$\therefore x \in (- \infty ,5) \cup (5, + \infty)$$

Range
 I can simplify the function to the form below, but I don't know how to go from there.

$$ f(x)= x + 5 + \frac {1}{x-5}$$
 

 

[Mark44]

Homework Statement:: Find the domain and range of following function: $f(x) =\frac {x^2 -1} {x-5}$.
Relevant Equations:: None

I can get the domain, but getting the range seems impossible.

Domain
$$x-5=0$$
$$x =5$$
$$\therefore x \in (- \infty ,5) \cup (5, + \infty)$$

Range
 I can simplify the function to the form below, but I don't know how to go from there.

$$ f(x)= x + 5 + \frac {1}{x-5}$$
 

For the equation $\frac{x^2 - 1}{x - 5} = y$, solve for x. When you do this, using the Quadratic Formula, the expression inside the radical must be greater than or equal to zero. That will give you the range; i.e., the possible values for y.

 

[vcsharp2003]

For the equation $\frac{x^2 - 1}{x - 5} = y$, solve for x. When you do this, using the Quadratic Formula, the expression inside the radical must be greater than or equal to zero. That will give you the range.

I get $y \le 10 - 4 \sqrt {6}$ or $y \ge 10 + 4 \sqrt 6$. I need to double check.

My initial reaction after looking at the simplified function form in my attempted answer was $y \in (-\infty, 10) \cup (10, + \infty)$

 

[Mark44]

I get $y \ge 10 - 4 \sqrt {6}$. I need to double check.

In my work involving the expression under the radical, I got this:
$(y - 10)^2 \ge 4\sqrt 6$

There are two sets that represent the range. What you showed isn't too far off, but isn't either of them, so yes, double-check your work.

My initial reaction after looking at the simplified function form in my attempted answer was $y \in (-\infty, 10) \cup (10, + \infty)$

No, that's not the range.

 

[FactChecker]

I get that it is equal to $x+5+\frac{24}{x-5}$.

 

[vcsharp2003]

In my work involving the expression under the radical, I got this:
$(y - 10)^2 \ge 4\sqrt 6$

There are two sets that represent the range. What you showed isn't too far off, but isn't either of them, so yes, double-check your work.

No, that's not the range.

It seems I have it correct now as mentioned in edited post#3.

 

[Mark44]

I get that it is equal to $x+5+\frac{24}{x-5}$.

I don't see how that helps in finding the range. This just shows that there is an oblique asymptote of y = x + 5.

 

[vcsharp2003]

I get that it is equal to $x+5+\frac{24}{x-5}$.

Yes, you're correct. My mistake in simplifying the original function.

 

[Mark44]

It seems I have it correct now as mentioned in edited post#3.

Yes, that's what I get. So the range is the set $\{y | y \le 10 - 4\sqrt 6 \cup y \ge 10 + 4\sqrt 6\}$.

 

[vcsharp2003]

Yes, that's what I get. So the range is the set $\{y | y \le 10 - 4\sqrt 6 \cup y \ge 10 + 4\sqrt 6\}$.

That was a great solution. Thank you for showing me the correct way to solving such difficult problems.

 

[vcsharp2003]

Yes, that's what I get. So the range is the set $\{y | y \le 10 - 4\sqrt 6 \cup y \ge 10 + 4\sqrt 6\}$.

If we had something like $f(x) = \frac {x-1}{x+2}$, then we couldn't use the approach mentioned in posted question. I am wondering how we could determine range in this case. I guess we would solve the equation $\frac {x-1}{x+2} = y$ for $x$ and then use the fact that $x \neq -2$ to get excluded values of $y$.

 

[FactChecker]

I don't see how that helps in finding the range. This just shows that there is an oblique asymptote of y = x + 5.

It changes the values of the function, so it changes the range.
The solution you got did not use this result (either the correct one or the incorrect one), so it was not relevant to your answer.

 

[Mark44]

It changes the values of the function, so it changes the range.
The solution you got did not use this result (either the correct one or the incorrect one), so it was not relevant to your answer.

I don't know what you mean by any of the above. Your work merely rewrites the given equation as a first-degree polynomial plus a rational function. All this does is to show the long-term behavior (oblique asymptote of y = x + 5) and a vertical asymptote (the line x = 5).

Yes, that's what I get. So the range is the set $\{y | y \le 10 - 4\sqrt 6 \cup y \ge 10 + 4\sqrt 6\}$.

The set above gives the range. One can verify this by looking at a graph on wolframalpha.

 

[FactChecker]

I don't know what you mean by any of the above.

If you had used the incorrect version to determine the range, it would have been wrong. You did not use this form at all, so it was not relevant to your answer.
Clearly, if you use the wrong function, you are likely to get the wrong range.

 

[Mark44]

If you had used the incorrect version to determine the range, it would have been wrong. You did not use this form at all, so it was not relevant to your answer.
Clearly, if you use the wrong function, you are likely to get the wrong range.

I still don't understand what you're trying to say.
The given problem is

Find the domain and range of following function: $f(x) =\frac {x^2 -1} {x-5}$.

If you're saying that I made a mistake somewhere, please show me what it was.

What I'm saying, is that $f(x) =\frac {x^2 -1} {x-5}$ and $f(x) = x + 5 + \frac {24}{x - 5}$ are equivalent. The latter form merely gives the oblique and vertical asymptotes, but not the range.

 

[FactChecker]

I still don't understand what you're trying to say.
The given problem is

If you're saying that I made a mistake somewhere, please show me what it was.

No. Everything you did was fine.
I wrote my reply based on the original hand-written post when I saw that the calculation was wrong. That hand-written version of the original was later removed and all the conversation was not shown in that tab. It wasn't until I got around to posting my correction that it said the post was removed and I went to the new post to put my correction in. I didn't notice all the intermediate posts that I had missed until later.

What I'm saying, is that $f(x) =\frac {x^2 -1} {x-5}$ and $f(x) = x + 5 + \frac {24}{x - 5}$ are equivalent. The latter form merely gives the oblique and vertical asymptotes, but not the range.

It is equal to the original function. If you use it to determine the range, it gives the range.

 

[berkeman]

I wrote my reply based on the original hand-written post when I saw that the calculation was wrong. That hand-written version of the original was later removed and all the conversation was not shown in that tab.

Yeah, the OP's original thread was deleted so that he could re-post using LaTeX. It sounds like you still had a copy of the original thread open for a while...

 

[Mark44]

Regarding your other question ...

If we had something like $f(x) = \frac {x-1}{x+2}$, then we couldn't use the approach mentioned in posted question. I am wondering how we could determine range in this case. I guess we would solve the equation $\frac {x-1}{x+2} = y$ for $x$ and then use the fact that $x \neq -2$ to get excluded values of $y$.

Yes, same technique to find the range: solve the equation $y = \frac{x - 1}{x + 2}$ for x. Aside from a vertical asymptote of x = -2, there is also a horizontal asymptote. That asymptote has an impact on the range.

 

[FactChecker]

Yeah, the OP's original thread was deleted so that he could re-post using LaTeX. It sounds like you still had a copy of the original thread open for a while...

Yes. I was really surprised that nobody had answered in all that time.
They were answering on the new thread. :-)

 

[vela]

I don't see how that helps in finding the range. This just shows that there is an oblique asymptote of y = x + 5.

I deduced the shape of the graph by seeing the function as the sum of $y=x+5$ and $y=24/(x-5)$, so there's a local minimum on the $x>5$ branch and a local maximum on the $x<5$ branch. Then I used calculus to find the local extrema.

 

[pasmith]

Write <br /> y = \frac{x^2 - 1}{x - 5} = \frac{(x - 5 + 5)^2 - 1}{x - 5} = x - 5 + \frac{24}{x-5} + 10. Now for x &gt; 5 set x = 5 + \sqrt{24}e^t = 5 + 2\sqrt{6}e^{t} to get <br /> \begin{split}<br /> y &amp;= 2\sqrt{6}(e^t + e^{-t}) + 10 \\<br /> &amp;=10 + 4\sqrt{6}\cosh t \\<br /> &amp;\geq 10 + 4\sqrt{6}. \end{split} For x &lt; 5 instead set x = 5 - 2\sqrt{6}e^t to get \begin{split}<br /> y &amp;= 10 - 4\sqrt{6}\cosh t \\<br /> &amp;\leq 10 - 4\sqrt{6}.\end{split}

 

[SammyS]

@pasmith
That's a very interesting parametrization .

 

[pasmith]

That's a very interesting parametrization .

The identities \left.\begin{split}<br /> |x| + a|x|^{-1} &amp;\equiv 2\sqrt{a}\cosh(\ln (|x|/\sqrt{a})), \\<br /> |x| - a|x|^{-1} &amp;\equiv 2\sqrt{a}\sinh(\ln (|x|/\sqrt{a})),<br /> \end{split}\right\} \quad a &gt; 0, x \neq 0 are occasionally useful.