# Can someone explain this to me?

#### retsuL

##### New member
Hi again math help community.

I was wondering how do you find the range for this kind of functions:

f(x)= x-(9/x), if x∈ (0 , 5(

To me the problem for this function is the second degree polynomial on the numerator.

x-(9/x) = ((x^2) -9)/x

#### Dr.Peterson

##### Elite Member
Hello. I'm not familiar with the notation (0, 5( but I'd guess it means all numbers in between 0 and 5. In the US, we would write that as interval notation (0,5). If the 5 were included, then we would write (0,5].
Though I've never seen this )a, b) notation, I'm pretty sure it means what we would mean by [a, b), and others would write as [a, b[. That is, a parenthesis facing inside is exclusive, and facing outside is inclusive. So (0, 5( means (0, 5].

As for the problem itself, my inclination is to sketch the graph of the function, and use that to determine the range. If necessary, the processes used for sketching will turn into proofs of various aspects (such as monotonicity).

#### MarkFL

##### Super Moderator
Staff member
We are given:

$$\displaystyle f(x)=x-\frac{9}{x}$$

Observing that $$x\ne0$$, then multiplying by $$x$$ and arranging in standard quadratic form, we find:

$$\displaystyle x^2-fx-9=0$$

The discriminant (which cannot be negative) is:

$$\displaystyle f^2+36\ge0$$

This implies the range of the function over its entire domain is all real numbers. However, the domain has been restricted. We should also notice that the given function has a vertical asymptote at $$x=0$$, and an oblique asymptote at $$y=x$$. We can see that as $$x$$ approaches 0 from the right, the function is decreasing without bound.

Now, if we consider:

$$\displaystyle f'(x)=1+\frac{9}{x^2}$$

This means for $$x\ne0$$ the function is strictly increasing, and for some positive number $$a$$, then for the domain $$(0,a]$$, the range of the function is then:

$$\displaystyle (-\infty,f(a))=\left(-\infty,a-\frac{9}{a}\right)$$

#### retsuL

##### New member
Hello. I'm not familiar with the notation (0, 5( but I'd guess it means all numbers in between 0 and 5. In the US, we would write that as interval notation (0,5). If the 5 were included, then we would write (0,5].

Is f defined for all numbers in between 0 and 5?

Can you find the limit of f, as x approaches 0 from the right?

Can you find the limit of f, as x approaches 5 from the left?

Also, please let us know if you've learned how to differentiate rational functions, like this one.

Before answering your questions, I would like to clarify that I am a precalculus student.
Is f defined for all numbers in between 0 and 5?
It is defined for all numbers the numbers in the given domain, except 0. The domain of f(x) is (0,5]

Can you find the limit of f, as x approaches 5 from the left?
No, since I'm at a precalculus stage I can not find limits neither differentiate functions.
I want to know if there is an analytic way(that does not include limits or differentiation) to justify the range for that domain.

#### retsuL

##### New member
As for the problem itself, my inclination is to sketch the graph of the function, and use that to determine the range. If necessary, the processes used for sketching will turn into proofs of various aspects (such as monotonicity).

I believe this is the approach I am looking for. I am not asking you to sketch the function it self, but may I ask for a demonstration?

#### retsuL

##### New member
We are given:

$$\displaystyle f(x)=x-\frac{9}{x}$$

Observing that $$x\ne0$$, then multiplying by $$x$$ and arranging in standard quadratic form, we find:

$$\displaystyle x^2-fx-9=0$$

The discriminant (which cannot be negative) is:

$$\displaystyle f^2+36\ge0$$

This implies the range of the function over its entire domain is all real numbers. However, the domain has been restricted. We should also notice that the given function has a vertical asymptote at $$x=0$$, and an oblique asymptote at $$y=x$$. We can see that as $$x$$ approaches 0 from the right, the function is decreasing without bound.

Now, if we consider:

$$\displaystyle f'(x)=1+\frac{9}{x^2}$$

This means for $$x\ne0$$ the function is strictly increasing, and for some positive number $$a$$, then for the domain $$(0,a]$$, the range of the function is then:

$$\displaystyle (-\infty,f(a))=\left(-\infty,a-\frac{9}{a}\right)$$
Although I don't have a deep understanding of the concepts of differentiation, I can tell you are using it to justify the monotony of the function and determine a maximum, right?
I must apologize since I did not give any other condition for this problem, as I already said I am a precalculus student, and I want to find the range with out differentiation.
This is still a useful answer since it is concise and accurate. It will help me for further studies of the subject.
Thanks a lot!

#### MarkFL

##### Super Moderator
Staff member
Although I don't have a deep understanding of the concepts of differentiation, I can tell you are using it to justify the monotony of the function and determine a maximum, right?
I must apologize since I did not give any other condition for this problem, as I already said I am a precalculus student, and I want to find the range with out differentiation.
This is still a useful answer since it is concise and accurate. It will help me for further studies of the subject.
Thanks a lot!
At the time of my posting, I was not aware that you are a PreCalculus student...I simply observed that this thread is posted in the calculus section and so assumed differentiation would be something allowed here.

As the derivative of the given function is always positive over its entire domain, we know that the curve is strictly increasing. But, we are only interested in part of the curve to the right of its vertical asymptote.

But, even without the calculus, armed only with the asymptotes and the $$x$$-intercepts $$(\pm3,0)$$ we can obtain a good picture of the function's behavior.

#### Dr.Peterson

##### Elite Member
I believe this is the approach I am looking for. I am not asking you to sketch the function it self, but may I ask for a demonstration?
The main graph sketching technique I expected you to use was differentiation. (If you don't want calculus used in an answer, don't post under calculus!)

But precalculus sketching techniques will give you the intercepts and where the function is positive or negative, and the asymptotes ("end behavior", which is really the same concept as limits). You would be mostly guessing about particulars, which you could partially support by plotting specific points.

But you can do more: Solve for x in terms of y using the quadratic formula, which will show that for any y, there are two x's, one positive and one negative, which implies that the branch for x>0 is one-to-one, and therefore monotonic. That is more than sufficient to answer the question.

But why was this problem assigned in a precalculus course? What techniques have you learned in this chapter, and were you given any examples of similar questions?

#### Otis

##### Senior Member
… The domain of f(x) is (0,5] …
Okay. That's not what you typed.

… I want to know if there is an analytic way …
Yes, there is.

Do you know that 9/x gets smaller, as x gets bigger?

What happens to 9/x, as x gets smaller? That is, as positive values of x approach zero?