Question

Need quick help plz :)

Answer 1

I’m trying to plugin values for x into the following function... plz check my zebra..\[Y = \left( 1+x \right)^{3/2}\]

Answer 2

I’m having a tough time drawing

Answer 3

Let me rearrange the function so we can plugin the x-values

Answer 4

I don't know what is wrong with your drawing pad, but in terms of graphing, you should try plugging in really large negative and positive values to get a sense for the long term behavior of the function.

Answer 5

Those large values might also help you determine if there are any asymptotes.

Answer 6

Keep going?

Answer 7

Glad its working for you now

Answer 8

Yes keep going, but remember what I told you as well

Answer 9

Let’s find out if I remember lol

Answer 10

Hold up...

Answer 11

You've changed the computation

Answer 12

I did? :# sorry ;-;

Answer 13

Be careful how you use parentheses.

Answer 14

Ok let me fix that in the net drawing

Answer 15

Parentheses always occur in pairs. You have five, but you only need four.

Answer 16

Parentheses open and close expressions. You have an expression that is being cubed. The expression being cubed needs to be enclosed with a set of parentheses. Then you've entered a value for \(x\) that also needs to be enclosed with a set of parentheses. No other parentheses are necessary

Answer 17

\(f(-2) = \sqrt{(1 + (-2))^3}\)

Answer 18

That's what you want

Answer 19

If we do it the way you posted above we get a different result

Answer 20

Because it looks like you're trying to cube the negative 2 only, but you have to add before you can cube

Answer 21

Looks right

Answer 22

Sooo how do we go about figuring out the domain and range of that thing

Answer 23

Keep it short tho ;-;

Answer 24

Well, once again we have a square root function so keep that in mind. You have to figure out what the lowest extreme is. Try to figure out the lowest negative value of x that is included in the domain. Remember that the square root of a negative number is imaginary. So what negative number can you put into x that will yield a positive value under the square root?

Answer 25

Once you figure out what the lowest negative number is for the domain, you'll know that the domain won't include any numbers that are more negative than the lowest negative number. Then from there, you can start trying numbers that are greater than the lowest negative number and work your way towards positive values. Then from there see if you can find any other restrictions on the domain.

Answer 26

Your goal for the domain is to find the restrictions (values excluded from the domain) and the extremes (upper and lower values of the domain)

Answer 27

Also, one more thing. It's probably a good idea to identify which kind of function is you're working with before even attempting to find the domain. I remember giving you a document that has the graphs of all the parent functions. Have you identified what kind of function you're graphing?

Answer 28

Yes, that. And be sure to look carefully. There are some things on there that may trip you up or lead you to become confused.

Answer 29

Wait, vertical asymptote has to do with domain right?

Answer 30

Knowing the vertical asymptote can assist you in determining the domain of a function.

Answer 31

Okie dokie

Answer 32

https://assets.questioncove.com/attachments/1560700647-5d065d7c4425911ac3399000-ParentFunctions.png

Answer 33

There's a direct link to the poster of parent functions. Save that link instead.

Answer 34

Thanks super man

Answer 35

Just realized that the value of x has to be either 0 or greater than 0 in order for the y-value to be a real number

Answer 36

Anything below 0...lets say -1 will make the entire root undefined

Answer 37

Yes, that's why I kept saying "Square root function"

Answer 38

Are you not familiar with how square roots operate?

Answer 39

So our domain is *x is the element of all real numbers, except x has to be either equal or bigger than 0*

Answer 40

Maybe 😆

Answer 41

Well, you still have to be careful. It is whatever that is under the square root that has to be greater than or equal to zero.

Answer 42

To find the domain you have to set whatever is under the square root greater than or equal to zero, then solve for x

Answer 43

Wait, what do you call this function tho??? \[Y = \left( 1+x ^{} \right)^{3/2}\]

Answer 44

Well, let me explain...

Answer 45

I literally tried a bunch of x values to realize that lol so yuhh x has to be greater then 0

Answer 46

Is it just a twin sister of square root function? Just rearranged like that?

Answer 47

There are a couple of ways to express the function. The way you have expressed it is the exponential form. A better way to express it is \(y = \sqrt{(1 + x)^3}\). In this way, you know that the expression \((1 + x)^3\) is under the square root. Therefore to find the domain you would set \((1 + x)^3 \ge 0\) then solve for \(x\).

Furthermore \(y = \sqrt{(1 + x)^3}\) is equivalent to \(y = (1 + x)^{3/2}\) because of the rule of exponents that says \(a^{1/n} = \sqrt[n]{x}\)

Answer 48

Hope that isn't too complicated or confusing for you

Answer 49

I literally know all that

Answer 50

So the domain is ....x has to be equal or greater than 0 right? Cuz any value less than 0 will result in undefined value for y

Answer 51

I explained how to FIND the domain. Please read my previous post again. There's something you're overlooking for some strange reason.

Answer 52

What tho?

Answer 53

Let me highlight it for you

Answer 54

Oh you mean the asymptote?

Answer 55

There are a couple of ways to express the function. The way you have expressed it is the exponential form. A better way to express it is \(y = \sqrt{(1 + x)^3}\). In this way, you know that the expression \((1 + x)^3\) is under the square root. Therefore

\(\boxed{\textbf{to find the domain you would set} (1 + x)^3 \ge 0\ \textbf{then solve for }x}\)

Furthermore \(y = \sqrt{(1 + x)^3}\) is equivalent to \(y = (1 + x)^{3/2}\) because of the rule of exponents that says \(a^{1/n} = \sqrt[n]{x}\)

Answer 56

Yes i did that already...wait let me draw.

Answer 57

Read the entire paragraph again so that you understand completely

Answer 58

If you did it already then you would know what the domain is

Answer 59

Nope that's not what I was asking you to do. If you were doing it the way that I asked, you would have started with this:

I clearly put in bold what to do

Answer 60

Eye see..

Answer 61

So you want me to only deal with what’s under the root...

Answer 62

It is clear that you don't know how to solve for x. You don't solve for x by plugging in values for x. You solve for x by ISOLATING x on one side of the inequality.

Answer 63

Sorry mate... let me try again..

Answer 64

How do i get rid of the freakin cube?

Answer 65

Cube root both sides

Answer 66

Very good and can you please write x and just x and avoid using the weird canadian symbology? It hurts my eyes and does not look like an x at all

Answer 67

Very good

Answer 68

That is the domain \(x \ge -1\)

Answer 69

Eye see what u did there...

Answer 70

What scenarios can i apply this in? Just the root functions?

Answer 71

Even root functions

Answer 72

Again, we can get this done much faster if u use simple instructions...and not math notations and stuff

Answer 73

It makes me panic lol

Answer 74

Certain things need to be explained first so that more confusion doesn't happen down the line.

Answer 75

K, the next question is very similar... i’ll See if i can apply this method on that one..k

Answer 76

One second, let me write errthing down... :)

Answer 77

I have to go tend to my own work now. You're on your own until probably midnight.

Answer 78

Okie dokie :)

Answer 79

Thanks for everything hero :)

Answer 80

yw

Answer 81

How do i find the range of this function again?

Answer 82

By testing values from the domain. Start with the lowest extreme value (x = -1). That will give you the lowest extreme value of the range. I have a question for you. What is the Upper extreme value of the domain?

Answer 83

Idk what upper extreme value is...

Answer 84

Tell me in simple words plz

Answer 85

If the domain of the function is \(x \ge -1\) what is the highest possible value of \(x\). What I'm trying to get from you is "What are the upper and lower limit values of the x axis? What is the lowest possible value of x. What is the highest possible value of x?

Answer 86

If x = -1 is the lowest extreme value, the highest could be any positive integer...

Answer 87

The upper extreme of the domain is positive infinity.

Answer 88

That’s what i was thinking but was too shy to say...

Answer 89

Cuz there’s no limit to the function ....it can go as high as it wants

Answer 90

And now you need to plot a few points to determine the direction of the graph.

Answer 91

Is the graph trending upwards or downwards?

Answer 92

Aaaa it doesn’t ask me to graph but it should go up...

Answer 93

As high as it wants to go... all the way to positive infinite

Answer 94

Let's back track for a minute. Did you find y when x = -1?

Answer 95

Whatever that y value is, that is the low end of the range.

Answer 96

The upper limit of the range is obviously infinity

Answer 97

Yes i did...it was 0

Answer 98

Okay great so the domain is \([-1,\infty)\) and the range is \([0, \infty)\)

Answer 99

Ah icy

Answer 100

So basically, when determining the range, you need to know three things: The direction of the graph, The lowest possible value of the range, the highest possible value of the range. You get those values by evaluating the function using the extreme values of the domain.

Answer 101

So there you go. The easy way to find the range.

Answer 102

Hope you're happy now.

Answer 103

Nevertheless, this method only works if the graph is continuous.

Answer 104

Thanks hero. :) you’re my avengers of maths :)

Answer 105

If there are any restrictions on x, then there will be more work involved because you have to account for asymptotes and such.

Answer 106

Yuhh...thank god i didnt have that for this question...

Answer 107

In this case the domain did have a restriction, but it is continuous on the domain \([1, \infty)\)

Answer 108

However, if it was not, you'd have to find the vertical asymptote and and exclude that value from the domain.

Answer 109

As for the domain, can i say like x has to be equal or bigger than -1?

Answer 110

I’ll keep that in mind :)

Answer 111

\(x \ge -1\) means the same thing as \([-1, \infty)\) Both are describing the domain.

Answer 112

Okie dokie

Answer 113

Okay, I'm not coming back until midnight. I was just taking a break.

Answer 114

Okay, thanks hero :)