Question

Someone help me find the domain and range of rational function.

Answer 1

Find the domain and range of the following ratuinal function.
\[F(x) = \frac{ 2 }{ x^2+1 }\]

Answer 2

I know i can’t set the denominator equal to 0 and solve for x in this case. What do i need to do to find the vertical asymptote?

Answer 3

@Hero

Answer 4

You still have to set the denominator equal to zero and solve for x because you have to prove that there is no vertical asymptote.

Answer 5

Let me draw and u evaluate

Answer 6

Wait, wouldn’t that give me zeros aka the two Points where the graph touches the x-axis?

Answer 7

Nope, it shouldn't. You should get imaginary numbers instead.

Answer 8

Imaginary numbers can't exist on a real number line.

Answer 9

How do i know whether i’ll Get real numbers or imaginary?

Answer 10

Solve this for \(x\):
\(x^2 + 1 = 0\)

Answer 11

I can’t, it’s a quadratic equation .... wait one second. Dad’s calling ...terribly sorry for the pause. Brb i promise.

Answer 12

Quadratic Equations can be solved for \(x\).

Answer 13

Sorry for the delay. Let me try to factor the quadratic equation.

Answer 14

What’s the rule when the denominator is not factorable?

Answer 15

You don't try to factor the denominator. You just set it equal to zero and solve for \(x\). There's no need to factor in this case

Answer 16

Damn, so does that mean I did it wrong? Ok let me set it equal to zero and see.

Answer 17

Just the denominator of course

Answer 18

While I set he denominator to zero, do I need to use the quadratic equation?

Answer 19

That's not necessary either.

Answer 20

Ok let me draw first and we’ll go from there...

Answer 21

Error = imaginary number right?

Answer 22

Well, you see, you can definitely take it a few steps further. Let me show you
$$\begin{aligned}
x^{2} & =-1\\
\sqrt{x^{2}} & =\sqrt{-1}\\
|x| & =i\\
x & =\pm i
\end{aligned}$$

Answer 23

Oh I didn’t know you could do that

Answer 24

Here we've shown that \(x\) is clearly an imaginary number which does not exist on the real number plane.

Answer 25

When the denominator is set to zero that is...

Answer 26

Otherwise, there are no restrictions on \(x\)

Answer 27

Hero, I don’t need to know much about imaginary numbers bro. I’m just looking for domain and range of the rational function. I just wanted to know if there are any restrictions in the domain or range that i need to include in the domain and range. Also, when the denominator resolves to be *imaginary number* that means there are no restrictions/vertical asymptotes?

Answer 28

Correct. If you set the denominator equal to zero and solve for x and get an imaginary number as the result, then that means no vertical asymptote exists for the rational function.

Answer 29

To find the range, find \(f^{-1}(x)\). In other words find the inverse of the function

Answer 30

Can you please do me a favour? Can you please find me a sheet that shows all the rules related to finding the vertical and horizontal asymptotes of rational functions?

Answer 31

There aren't really that much rules.
To find the vertical asymptote,
set the denominator equal to zero, then solve for the variable.

To find the range, find the inverse of the function. The domain of the inverse of \(f\) is the range of \(f\).

Answer 32

Oh sorry, you wanted horizontal asymptote. Hang on...

Answer 33

One second let me fix a typo on that.

Answer 34

If the DEGREE of the leading terms of both the numerator and denominator are equal, then you have to consider the long range behavior. In the case, in the long term, the leading terms will dominate and the horizontal asymptote will approach the leading coefficient of the numerator over the leading coefficient of the denominator

Answer 35

If the DEGREE of the leading term of the numerator is SMALLER than the DEGREE of the leading term in the denominator, then the horizontal asymptote will be y=0

Answer 36

Typing too fast sorry. Had to fix another typo.

Answer 37

If the DEGREE of the leading term of the numerator is BIGGER than the DEGREE of the leading term in the denominator, then you have to perform long division to find the asymptote.

Answer 38

The asymptote will be the quotient of the divison

Answer 39

And that's it

Answer 40

I thought when the numberator and the denominator are equal, you divide the leading terms out. What do you mean by *you have to consider long division?*

Answer 41

You're mixing things up.

Answer 42

I mentioned the coefficients of the leading terms. Yes the variables will divide out leaving the coefficients.

Answer 43

For example....

Answer 44

\(f(x) = \dfrac{2x^2}{3x^2}\)

The long term behavior of \(f(x)\) is that as \(x \to \infty\) f(x) approaches \(\dfrac{2}{3}\)

Answer 45

Which is the horizontal asymptote

Answer 46

Yes that’s right. And that’s how i find the horizontal asymptote

Answer 47

Bro, I appreciate your work but can you please find me a picture that i can save on my ipad?

Answer 48

With all the rules to find the vertical and horizontal asymptotes of rational functions thanks

Answer 49

I just gave them to you. You can save what is above.

Answer 50

It’s hard to save that as a picture. It’s gonna take like 4 screenshots to capture everything

Answer 51

I got you. Hang on

Answer 52

Thanks bro. Let’s move on to finding the horizontal asymptotes of that function.

Answer 53

If the DEGREE of the leading terms of both the numerator and denominator are equal, then you have to consider the long range behavior. In the case, in the long term, the leading terms will dominate and the horizontal asymptote will approach the leading coefficient of the numerator over the leading coefficient of the denominator

If the DEGREE of the leading term of the numerator is SMALLER than the DEGREE of the leading term in the denominator, then the horizontal asymptote will be y=0

If the DEGREE of the leading term of the numerator is BIGGER than the DEGREE of the leading term in the denominator, then you have to perform long division to find the asymptote.

Answer 54

got it. Let’s find the inverse of the function now.

Answer 55

Wait, you didn’t including the *setting the denominator to x* to find the horizontal asymptote

Answer 56

...of a quadratic function.

Answer 57

You lost me there. I explained the three cases for finding the horizontal asymptote

Answer 58

"Setting the denominator equal to Zero and solving for x" applies when finding the vertical asymptote.

Answer 59

Omg sorry /.-

Answer 60

Let’s figure out the inverse of that function and finish this question.

Answer 61

To find the inverse, First replace \(f(x)\) with \(y\)

Answer 62

Next swap the \(x\)'s and \(y\)'s

Answer 63

Okay multiply both sides by the denominator

Answer 64

What do i do next?

Answer 65

Simplify the RHS by cancelling

Answer 66

RHS = Right Hand Side of the equation

Answer 67

Now divide both sides by \(x\)

Answer 68

Subtract 1 from both sides. By now you should be able to figure out what we are actually doing.

Answer 69

So we basically switched the variables and solved for y? Is that what we did

Answer 70

Yes, we're solving for \(y\). Finish doing so

Answer 71

Okay, replace the 1 with x/x

Answer 72

Sorry I didn’t get what x/x mean

Answer 73

x over x

Answer 74

Combine the fractions under the root

Answer 75

Do you know how to combine fractions?

Answer 76

Nope

Answer 77

\(\dfrac{a}{b} + \dfrac{c}{b} = \dfrac{a + c}{b}\)

Answer 78

Take a common denominator

Answer 79

Omg lol

Answer 80

What do i do next?

Answer 81

Find the domain of \(x\)

Answer 82

I thought we were looking for the range

Answer 83

We are. Remember, I said earlier that the domain of the inverse is the range of the original function.

Answer 84

We had all real numbers for domain

Answer 85

That was for the original function. We need the domain of the inverse now

Answer 86

Jesus :/ how do we get that?

Answer 87

Well, first do you notice any restrictions on the domain?

Answer 88

Original function? Nope

Answer 89

On the domain of the INVERSE. We're focused on the inverse right now my friend.

Answer 90

You spent all that work and time finding the inverse. Now we will deal with that only for a bit

Answer 91

Oh my bad...yes I do see a restriction

Answer 92

How many restrictions do you see?

Answer 93

I can’t plug-in 2 cuz that’d make the sq root undef

Answer 94

I can’t plug-in any value less than 2...I think

Answer 95

Meaning whatever value i plug-in has to be greater than 2

Answer 96

Actually, here's the way you should see it. There's an \(x\) in the denominator of the fraction so we know \(x \ne 0\)

Answer 97

Yes that too

Answer 98

And actually \(x\) can equal 2. We just can't have any values greater than 2

Answer 99

What happens if \(x > 2\)?

Answer 100

It becomes negative in the numerator and that’s a no no to find the sq root

Answer 101

The square root can't be negative otherwise we get imaginary numbers. So we know that \(x\ne 0\) and \(x \ngeq 2\)

Answer 102

So what is the domain of the inverse here?

Answer 103

All real numbers except x can’t be 0 ad 2

Answer 104

Nope

Answer 105

Can x be less than zero?

Answer 106

Can x be greater than 2?

Answer 107

No fr both

Answer 108

Okay, so what is the domain?

Answer 109

X has to be greater than 0 and x has to be greater than 2

Answer 110

Negative. x cannot be greater than 2. Why don't you try evaluating a number greater than 2 and see what happens.

Answer 111

It’ll make the numerator negative and ultimately the everything in the sq root

Answer 112

Okay, so last try. What is the domain of the inverse?

Answer 113

Let me think...one second

Answer 114

X has to be less than 2, and x can’t be equal to 0

Answer 115

Which means the domain of the inverse exists on what interval?

Answer 116

BTW, I'll say it again, \(x\) can be 2

Answer 117

I meant less than or equal to 2

Answer 118

If you just say \(x \le 2\) then you're also including negative numbers which we already said are not possible here. The domain of the inverse is \(0 < x \le 2\)

Answer 119

😮 I forgot about that..yeah

Answer 120

I was just focusing on 2 there

Answer 121

This is what the graph looks like: https://www.geogebra.org/classic/mnzmgbqx

Answer 122

Wait hang on. I just realized that for the original function, you have to write it as \(0 < f(x) \le 2\) and no you cannot write it as you wrote it.

Answer 123

Why fx?

Answer 124

Because the range is the set of y values

Answer 125

Cuz it’s the domain of the original function?

Answer 126

I meant range

Answer 127

Exactly

Answer 128

Why’s there an x

Answer 129

I corrected that by writing f(x) instead. \(\color{#0cbb34}{\text{Originally Posted by}}\) @Hero
Wait hang on. I just realized that for the original function, you have to write it as \(0 < f(x) \le 2\) and no you cannot write it as you wrote it.
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 130

Did you not see that?

Answer 131

It's easy to confuse one thing with the other because we were talking about the domain of the inverse being the range of the original function. You have to concentrate to keep up with the variables that differentiate both sets.

Answer 132

Sometimes I slip up and say one thing when I mean the other. It's very easy to slip up in this case.

Answer 133

NO.

Answer 134

Let me delete the wrong stuff so you're no longer confused.

Answer 135

So the RANGE of the original function is \(0 < f(x) \le 2\)

Answer 136

Okay?

Answer 137

Yes thanks for your help