Question

Find the domain and range of the following function.

Answer 1

\[F(x) = \frac{ 1 }{ (7-x)(x^2-1) }\]

Answer 2

For domain, the denominator cannot equal 0

So do
\((7-x) = 0\) and \((x^2-1)=0\)

This will tell you what cannot be part of the domain

Answer 3

I see what you did. Let me try brb

Answer 4

If you need help solving for x, ask

Answer 5

@hero i need his help. He’s the best. No offence tho :)

Answer 6

So let me first ask you: what is the domain & range? and what is a discontinuity?

Answer 7

It's best that you understand the significance of what you're doing so then you can attack any question in any method and see the purpose behind it

Answer 8

Domain the is relationship between independent variables and dependent variables. Domain shows if there’s a point y on the graph for every x value

Answer 9

in layman's, it's just the set of x values in a function. range is the set of y values, which are dependent on the domain values.

Answer 10

Ok now how do we get the domain and range of that function?

Answer 11

Why can’t we expand the denominator and find the zeros?

Answer 12

It's good to always know the end behavior of various functions (trig, trig inverse, power, exponential, etc) so you can visualize it in your mind how you should approach.

For now, we can find the domain. Let's start with assuming that x exists for all real values. Now, we have to look for discontinuities.

Do you know what is a discontinuity, and what discontinuity is likely in this function?

Answer 13

Yes vertical and horizontal asymptotes where the curve of the graph approaches the restriction/asymptotes but never quite intersects

Answer 14

Yeah. So the vertical asymptote is a restriction on the domain, and the horizontal is a restriction on the range, correct?

Answer 15

Bro, let’s speed this up, i have like 9 more freaky questions like these

Answer 16

Yes, i know those things.

Answer 17

okay haha.

So to find the vertical asymptote set the denominator to 0 and solve for x using the zero product property.

Answer 18

Did you get that?

Answer 19

Yes sir, let me draw ...

Answer 20

Don’t give me hints. Let me draw first

Answer 21

Sure sure. No need to draw/graph though, but if it helps you visualize, then it's great.

Answer 22

Shoot, my pencil is out of juice. One second, let me charge it. Thanks

Answer 23

I mean draw out the answer

Answer 24

ohhh. yeah ok

Answer 25

check your second equation again. You brought the -1 over but forgot to change the sign.

Answer 26

x^2 - 1 = 0
+1 +1
x^2 = 1
so...

Answer 27

Close. is that the only root of √1 ?

Answer 28

So my pen was freaking out while i was drawing. Sorry

Answer 29

What do i do next?

Answer 30

There we go. So you got x=7, -1, and 1. So those are your vertical asymptotes

Answer 31

That’s it? :O

Answer 32

Yeah. So your domain is X exists for all real numbers, x≠7, -1, 1.

Answer 33

Alright, let’s find the horizontal asymptotes now :)

Answer 34

Now for your range, recall that the horizontal asymptote is a restriction on the range.
So to find the horizontal asymptote, do you know the rules?

Answer 35

Sorry bro, we have a thunderstorm here, it knocked out lights out for a bit. But yeah, i can’t recall the rule tho. Plz hep

Answer 36

Omg i actually do

Answer 37

is it hurricane dorian?

rules:
1. if the degree of the num < degree of denom, then y=0 is the horizontal asym.
2. if degree of num > degree of denom, it is a oblique asymptote (no horizontal, use synthetic devision to find it)
3. if degree of num = degree of denom, divide leading coefficients.

Answer 38

If the leading degrees are equal, you divide leading coefficients
Second rule is, if the degree in the numerator is bigger than the degree in the denominator, there’s no horizontal asy
Finally, if the degree in the denominator is bigger than the numerator, the horizontal asym is y=0

Answer 39

yeah

Answer 40

Do you really think i’d Be sitting here if we had hurricane Dorian ripping through our house? Lol

Answer 41

Yeah I was being sarcastic lmao
but the Fla side of it isn't bad

Answer 42

Ok so we have a 1/ (7-x)(x^2-1). The degree in the denominator is bigger than numerators...so horizontal asym is y = 0

Answer 43

yeah that's it

Answer 44

Really? Damn, that was easy lol

Answer 45

Let me write everything down...brb thanks tho :)

Answer 46

np

Answer 47

Wait, hold on, why didn’t i get y=0 in the last question tho. The last question was y = 2/x^2+1

Answer 48

Well it should be y=0. You could prove it by plugging in the limit of x approaching to positive/negative infinity.

Answer 49

Ok so when i try to find the vertical asymptote, i got imaginary number right.

Answer 50

Yeah. So that just means that there is no vertical asymptote.

Answer 51

If you solve for a vertical asymptote and you end with a complex solution, it means that there is no vertical asymptote. You could check by graphing or using limits.

Answer 52

Then i try to find the horizontal asymptote, but it turned out i had to find the inverse of that function since the domain of the inverse is the range of the original function. So i have rewrite the equation in terms of x then isolated for y again and then simplified and did a combine fractioning. Then i tried to see what x values would make the sq root undef...and lastly, wrote the domain of the inverse function in terms of range of the original function....it was a pain.. x.x

Answer 53

Yeah there was no need for that

Answer 54

You can later go in my profile and see the last question. It’s pretty long tho.

Answer 55

Ok let me finish this and i’ll Be right back :) thanks

Answer 56

np

Answer 57

What’s the difference when you write 1^0 and 1^1? They both equate to be 1 anyway.

Answer 58

That's something to ask the international congress of mathematicians because powers to 0 and 1 are commonly debated.

Since the base is 1 they just somehow equate to the same. There's no difference i suppose other than the exponent itself

Answer 59

Cuz the thing is, i’m Trying to count the power numerator and the denominator and idk what to do there since the denominator is in parentheses

Answer 60

Just compare the degrees, there's no reason to go any further than that. but what question is it?

Answer 61

Same question...lol

Answer 62

the one we were just on? just always consider the exponent to the first power, because if it were any other whole number, you'd consider it the same.

Answer 63

Got it

Answer 64

So this is the range i got.

Answer 65

yeah correct

Answer 66

K next question. The aim is to find the domain and range of the following function. \[F(x) = \sqrt{x+3}\]

Answer 67

What do you call this type of functions again?

Answer 68

Square root functions

Answer 69

I see. Ok so how do we go about finding the domain?

Answer 70

We know that i can’t plugin any negative values into the sq root function.

Answer 71

So this is why I mentioned it's important to know very well the behavior of several functions.

So here, the part under the radicand must be greater than or equal to 0, as you know. There can't be any complex things graphed on a cartesian plane. So set the part inside the radicand to ≤0, solve for x. You'll get your domain.

Answer 72

We also know that w/e number we plugin, must not make the resultant integer negative.

Answer 73

Square root functions have lower bound, no upper bound, and there is no sign of vertical transformation, so you can already infer the range from that.

Answer 74

I think the domain is all real numbers except, x>0

Answer 75

Actually wait... omg /.- i’m So stupiff

Answer 76

Range* and it's x≥0

Answer 77

All real numbers except, x has to be equal to or greater than 3

Answer 78

Right

Answer 79

Can you please use simple math language, i’m Having hard time trying to understand what you’re trying to say about the range of the root func

Answer 80

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Ferredoxin4
Square root functions have lower bound, no upper bound, and there is no sign of vertical transformation, so you can already infer the range from that.
\(\color{#0cbb34}{\text{End of Quote}}\)
Like i dont get this lol

Answer 81

ok yeah I got you. look at this example

Answer 82

maybe @hero can take over

Answer 83

Is the last graph *bounded both above and below*?

Answer 84

yeah

Answer 85

I see. Thanks bro. Really appreacted your help :)

Answer 86

Great job explaining @Ferredoxin4

Answer 87

Thank you (:

Answer 88

How do i find the range of the root function again?

Answer 89

Can someone hep?

Answer 90

Is there a method to finding the range of the root func?

Answer 91

Ferredoxin explained it already, the function is bounded in the bottom. Naturally the range of the square root function is y≥0

Answer 92

The method is to use limits. Otherwise know the behavior of those functions like Ferredoxin said and apply the transformations

Answer 93

K got it. Thanks.