Question

Math help plz

Answer 1

Find the domain and range of the following function.
\[F(x) = (x-2)^{2/3}\]

Answer 2

@Ferredoxin4

Answer 3

Please tell me i don’t have to take the inverse of that monster. :/

Answer 4

dude just take the limit of x approaching 2 and you'll find a critical point.

Answer 5

Let me draw the solution and u check

Answer 6

ok

Answer 7

It doesn’t ask me to use limits here

Answer 8

Try all those and use those limit formulas to help you out in the future as well

Answer 9

I don't really know a way to do it without limits. If your teacher tells you not to use limits then you have a bad teacher because he/she making you do more work xD

Answer 10

Trust me when I do limits, i’ll Come back to these questions and take their souls

Answer 11

Another idea. Look for restrictions. Let me see myself

Answer 12

With limits all this would be a breeze

Answer 13

Is my inverse correct tho?

Answer 14

lemme see

Answer 15

Now that I have the inverse, i’m Going to look for restrictions now

Answer 16

yes it's right

Answer 17

I feel like I deserve a medal for not making any mistakes in my solution.

Answer 18

Now I’m going to plugging a few values of x and see what are the restrictions

Answer 19

Ok so i’ve Realized that x has to be equal or greater than 0 to have real numbers...anything below 0 will give undefined values

Answer 20

Thus, domain is all real numbers except x has to be equal or greater than 0

Answer 21

Yeah true.

I'll give you a suggestion if they allow it so you don't need to do a lot of work. Use mathematical induction.

It's a root function, so try setting the terms inside the radicand to 0 and see what x can't equal. You'll get your domain. Then do the similar with range.

Answer 22

Can you use mathematical induction?

Answer 23

Nope

Answer 24

Can I use this method for both even and odd root functions?

Answer 25

Yeah what matters more here is the root in any case.

Answer 26

When u say root, what do u mean?

Answer 27

√

Answer 28

I see

Answer 29

I’ve worked way to hard to learn this method...don’t lure me tho.

Answer 30

Let’s do the range now

Answer 31

Since we’ve found the domain of the inverse func, it’s the range of the original function

Answer 32

yeah

Answer 33

So the range is ... y can be any real numbers except, it has to be bigger or equal to 0

Answer 34

yes

Answer 35

domain is all real numbers

Answer 36

So to figure out the domain, do i plugin x values from the range back into the original function to find the domain?

Answer 37

I mean I just used limits here but yeah you could do that but just note it won't always work, it's just a nice strategy in understanding because of the Intermediate Value Theorem and understanding continuity.

Answer 38

1. Use Limits
2. Graph
3. Do it old school. Plug in values near critical points and know the end behavior, boundness, and restrictions of all functions. Notice function transformations.

Answer 39

That's the approach to domain and range

Answer 40

/I’ll do that when i reach limits in my textbook. But for now, how to i get domain algebraically

Answer 41

You just have to spot the constraints

Answer 42

Let me ask you: How does your textbook approach this?

Answer 43

Let me do one sample example from my textbook...they literally did it in two lines. No solution no nothing...

Answer 44

sorry, i meant 0<n<1 *

Answer 45

maybe @Hero knows a bit more to these questions, he can take over

Answer 46

If \[G(x) = \sqrt{7-x} \], then, for an even root to be defined, the expression under the root must be nonnegative: \[7-x \ge 0, 7\ge x.\] Thus, \[Dom {G} = (-\infty ,7], \] or \[Dom {G} = {x \epsilon \mathbb{R} : x \le 7}\]

Answer 47

This is literally how they do examples in 1 or 2 straight lines...

Answer 48

Yeah so that's actually what you need to do here. They are using end behavior to solve.

Answer 49

Odd radicals are Dom=xER, even radicals are always dom=xER: x≥0.

Answer 50

I c

Answer 51

Convert it to radical form. Forget what's inside the radicand, look at the odd or even root. It's odd (3), so it is automatically all real numbers for domain.

Answer 52

They are just doing a transformation by squaring it. Don't get fooled by that.

Answer 53

Can you explain what’s *transformation by squaring it* means?

Answer 54

\[\sqrt[3]{(x-2)^2}\]

The term that is squared is just a transformation to the parent function, \[\sqrt[3]{x}\]

Answer 55

Oh that transformation...i see now .

Answer 56

It's throwing you off. Pay attention to the parent function, not the radicand.

Answer 57

Got it

Answer 58

First check if it is odd or even, then move on accordingly. Odd will finish your task easier for domain, but if it's even, you must set the radicand to 0 and see where x≥n

Answer 59

Can you get me a chart of all the parent functions there are in our galaxy? Thanks bro

Answer 60

I see

Answer 61

Bless ur soul buddy

Answer 62

There's also trig and trig inverse functions, you can find that on your own i suppose.

Answer 63

K so the thing is...there’s other rational functions that look really scary..

Answer 64

How do i find the domain and range of this beast?

Answer 65

Do i have to find the vertical asymptote first?

Answer 66

I know that the value of x in the denominator has to be equal to or bigger than 2

Answer 67

To find the vertical asymptote, you would have to set the expression under the radical greater than or equal to zero, but in this case the expression can't equal zero either so use greater than zero instead. So set \(x - 1 > 0\)

Answer 68

Or rather that is to find the domain.

Answer 69

We can’t plugin the value of x = 1 in the denominator because that’d make the entire denominator equal to 0

Answer 70

So that is your asymptote. Now use \(x - 1 > 0\) to find the domain

Answer 71

Do you know the difference between \( >\) and \(\ge\) ?

Answer 72

Ofc

Answer 73

If you're using \(\ge\) you're saying that x can equal 1 which it cannot

Answer 74

Oops That was a mistake

Answer 75

So that is your domain

Answer 76

What bout range

Answer 77

One second

Answer 78

The power in the denominator is smaller than the power of numerator....so no horizontal asymptotes

Answer 79

That is correct

Answer 80

Nise

Answer 81

So what you want to do to find the range, is to find the lower bound. To do that you have to try values of \(x\). Try x = 2 to find the lower bound

Answer 82

Let’s first pray to the anime god to keep my internet alive so i can finish this question in my lifetime

Answer 83

Is that correct?

Answer 84

It is correct

Answer 85

What do i do next?

Answer 86

And the upper bound is ??

Answer 87

Infinite

Answer 88

The process is always the same. Finding the upper and lower bounds

Answer 89

To find the upper bound, you have to know which direction the graph of f(x) is going and whether there is an upper bound. Infinity means that f(x) continues to move in a positive direction and has no upper bound.

Answer 90

Wait, are you telling me to memorize the parent function of the transformed function ???

Answer 91

Infinity is what you put in place of an upper bound and just means that there is no upper bound. It means that the function continues in the positive direction forever

Answer 92

...the behaviour our parent and its transformed function?

Answer 93

The true way to find the upper bound is to apply the limit

Answer 94

Yes, that’s precisely why i said infinite because i thought the curve of our transformed function would go on for ever

Answer 95

I’m not at limit’s chapter mate. ...

Answer 96

So how do i fine the upper bound of our domain ?

Answer 97

I mean range

Answer 98

You don't have to use limits, but if I were you I would learn it so that you can justify what is happening as x gets larger and larger. For now, you can just use values such as x = 10, 100, 1000, 10000, 100000

Answer 99

You have to justify using infinity in place of the upper bound. That is, if you are not allowed to use a graphing utility to observe the graph.

Answer 100

That’s true, i’m Not allowed to use a graphing Calc in our class.

Answer 101

I’ll lurn limits when i get to the limits chapter

Answer 102

Help, then use that method. Keep using bigger and bigger values of x in your function and find the value of f(x) for each value of x to justify infinity since you can't graph or use limits.

Answer 103

I’ll comeback and do these questions all over again...

Answer 104

Once you find the value for x = 1,000,000, then you end by saying, As x gets larger and larger, f(x) approaches infinity

Answer 105

So the or rather in this case you're demonstrating that x has no upper bound so the domain is \((1, \infty)\)

Answer 106

I’ll just look at the parent function and it’s transformation and make my judgements from there. I ain’t going to manually plugging a million x values into the function just to find their f(x)

Answer 107

It's just seven computations. But technically that is the correct way to do it. That function is not on any of those parent function graphs. There's no function that has a square root in the denominator.

Answer 108

Hero. I have a major question for you....

Answer 109

A function like the following... what’s the parent function of that bro??

Answer 110

What’s their parent functions look like?

Answer 111

That's what I'm trying to tell you. There's no parent function for every function. Some of them you have to do tedious work to find domain and range

Answer 112

......:/

Answer 113

However, here's the thing. All you have to do is just identify the restrictions.

Answer 114

What’s the shortest way to solve these questions? Limits?

Answer 115

In the case of \(f(x) = \dfrac{\sqrt{x-2}}{(x -5)(x + 2)}\) you know that there are two values in the denominator that \(x\) cannot equal. In the numerator you know that \(x - 2 \ge 0\)

Answer 116

So you can use that information to help you figure out the domain.

Answer 117

Set each parentheses equal to zero and solve for x? That’d be my asymptotes

Answer 118

Correct. The asymptotes will tell you at least two values are excluded from the domain. Meanwhile, \(x - 2 \ge 0\) will tell you the rest of the values that are excluded from the domain and all of that together will tell you what the domain is.

Answer 119

I see....and lastly...how do i find the domain of the following function...

Answer 120

I have the inverse..

Answer 121

For that you should ask yourself if there is anything that is restricting x i.e \(x\) being in the denominator or restricted by a root.

Answer 122

One thing at a time bro.

Answer 123

We only use the inverse for finding the range

Answer 124

Find the domain before you find the range

Answer 125

We don't need the inverse to find the domain of the original.

Answer 126

That may confuse you though

Answer 127

Let’s not confuse me...

Answer 128

So i’ve Figured out the range of that function... and it’s y can be any real number except y has to be equal to or bigger than 0

Answer 129

Use \(f(x) = (x - 2)^{2/3}\) to find the domain
Use \(f^{-1}(x)\) (the inverse) to help find the range.

Answer 130

Have you found the domain yet?

Answer 131

I have the range of that function but no domain :/ that’s what i need your help for

Answer 132

The order in which to find these sets are to find the domain BEFORE finding the range.

Answer 133

You mean domain of the inverse? Yes

Answer 134

Nope, I mean the domain of the original function. I see that you are really confused now.

Answer 135

We've always found the domain or the original function first before attempting to find the range.

Answer 136

You're trying to do something different here.

Answer 137

Wait, let’s stop. Let me draw the answer and we’ll go from there.

Answer 138

I'm trying to help you avoid disaster if you take a test. Attempting to find the range first before the domain is mathematically illogical

Answer 139

We need to use \(f(x) = (x - 2)^{2/3}\) to find the domain first.

Answer 140

Ok let’s do that

Answer 141

Okay, so what are the restrictions on \(x\) if any?

Answer 142

Is there any x value you can put into the function that will lead to an undefined output?

Answer 143

Umm... let me check.. one sec

Answer 144

Whenever there is a restriction on \(x\) what are the usual conditions?

Answer 145

Bro I think there’s something wrong with my calculator... I’m getting 1 for pretty much every x value...

Answer 146

I don’t really have a power button that can do a power of 2/3 so I divide 2/3 and use the decimal version of the 2/3 power to solve for the x but i’m Getting 1 for everything over here

Answer 147

Can you square the value then apply the third root?

Answer 148

So negative x values r giving me errors..

Answer 149

Take a screenshot of your calculator and upload it here

Answer 150

I want to see what kind of calculator you're working with

Answer 151

You don’t wanna see my calc

Answer 152

I do actually

Answer 153

No u dont

Answer 154

Will you just upload it bro. No one is going to judge the calculator

Answer 155

Let me tell u the name it’s Casio fx-260 solar

Answer 156

It’s a monster

Answer 157

You have an \(a^{b/c}\) function. You need to learn how to use that

Answer 158

Let me see if I can find an emulator

Answer 159

I see it

Answer 160

So it makes fractions

Answer 161

Correct

Answer 162

Exponential fractions

Answer 163

If you think you've figured it out then evaluate f(3) and f(10) let me know what you get

Answer 164

Hold up..I need u to check if I did this right... 4-2 raised to the power of 2/3 i got 1.58

Answer 165

bro, evaluate f(3) and f(10)

Answer 166

Ok

Answer 167

For f(10) I got 3.65

Answer 168

Nope, that is not correct

Answer 169

You're doing something different from what you did the first time

Answer 170

Make sure you are entering \((10 - 2)^{2/3}\)

Answer 171

Just to be on the same page we r doing this right?

Answer 172

Yes, you can actually evaluate that manually

Answer 173

and if you can do it manually and get the result, the calculator should match what you get manually

Answer 174

4

Answer 175

Boom!

Answer 176

Some of your woes are resolved

Answer 177

I wasn’t using the brackets

Answer 178

Yep, you gotta use them

Answer 179

What’s woes?

Answer 180

Not knowing how to evaluate functions with exponential fractions

Answer 181

or rather fractional exponents

Answer 182

on your calculator that is

Answer 183

Anyways, where were we

Answer 184

K let’s focus...

Answer 185

Restrictions

Answer 186

Right find a restriction on \(x\). My question to you was whenever there are restrictions what are the usual conditions?

Answer 187

Let me check if there r any restrictions in the domain

Answer 188

What do u mean by usual conditions?

Answer 189

X can be either smaller, bigger or equal to n

Answer 190

I'll just give you this one. Usually when there are restrictions, x is either in the denominator of a fraction or under a root.

Answer 191

If \(x\) isn't restricted by an even root or a denominator then there is no restriction

Answer 192

Which means what for the domain if there is no restriction?

Answer 193

There r no asymptotes or graph where x doesn’t have a value for fx

Answer 194

So ultimately what does that mean for the lower and upper bound of the domain?

Answer 195

If there is no restriction on x

Answer 196

If there r no restrictions that means there’s a value for y for every x value

Answer 197

There’s no discontinuity

Answer 198

Am I getting this wrong?

Answer 199

How would you express the domain for a function with no restrictions on \(x\)

Answer 200

Omg.... all real numbers

Answer 201

Thank you

Answer 202

I’m a simple man, ask me in simple terms

Answer 203

Mathematically that is equivalent to writing \((-\infty, \infty)\)

Answer 204

Now we can find the range

Answer 205

Sure

Answer 206

I gave u the inverse

Answer 207

One second ...

Answer 208

I really wanna sleep tonight tho

Answer 209

Can we hurry this?

Answer 210

Okay, so you found that \(f^{-1}(x) = \sqrt{x^3} + 2\)

Answer 211

Tell me, how do we find the restrictions on \(x\) in this case?

Answer 212

Plugin a few points and see what makes the function undefined

Answer 213

X has to be equal to or bigger than 0

Answer 214

Have you not been taught to set the expression under the radical greater than or equal to zero?

In other words, have you set \(x^3 \ge 0\) ?

Answer 215

In order to find the domain?

Answer 216

I haven’t touched math in 10 year

Answer 217

I’ve already tested the x values ....

Answer 218

Bro, we just did several of these problems involving finding a domain and anytime you have an expression under an even root you're supposed to set it greater than or equal to zero to find the domain. We did at least five problems like this. I was giving you time to figure it out for yourself

Answer 219

I’ve a,ready done that part bro

Answer 220

What do you mean you've already done it?

Answer 221

I’ve tried a few values of x in the inverse function..

Answer 222

I don't want you to do it that way. I want you to solve this \(x^3 \ge 0\)

Answer 223

Oh is it bcuz it’s an even root function....omg /.- ok lets draw this baby

Answer 224

So I was right..x has to be equal or greater than zero

Answer 225

Well actually we have to replace x with y then it is correct.

Answer 226

The range is \(y \ge 0\)

Answer 227

We did it.

Answer 228

🎉

Answer 229

I mean finding the range is confusing for you if we use the inverse method.

Answer 230

The only thing that’s eating my soul is to remember how to approach an even or odd root of inverse function...that’s something I have to remember from here on

Answer 231

I think there is a reason most avoid the inverse method.

Answer 232

Apart from that...finding the inverse of a original function is no problem for me

Answer 233

I’m going to make notes with enough detail on this topic that even an ape will be able to do it

Answer 234

Anyways...appreciated ur help. God bless u

Answer 235

Have a good night...I’m going to finish making notes here

Answer 236

Ciao

Answer 237

Here's a resource you can study: https://www.purplemath.com/modules/invrsfcn3.htm

Answer 238

Thanks bud.

Answer 239

But there doesn't seem to be a resource for fractional exponent expressions

Answer 240

That's the only draw back however

Answer 241

It’s ok

Answer 242

If you have something like \(f(x)= (x - 1)^{3/2}\) if you re-write the function as \(f(x) = \sqrt[2]{(x - 1)^3}\) then you will see that the \(x\) is not restricted but the range is restricted.

Answer 243

Because of the square root

Answer 244

Okay, that was confusing. I was wrong about that

Answer 245

The domain is restricted and the range is restricted as well

Answer 246

Basically if you have an even root you are going to have restrictions in the domain AND the range

Answer 247

But it’s less work tho

Answer 248

If it’s a radicand function, i can just take w/e is under the radicand and set that to inequality and go from there..

Answer 249

Anyways.. i wanna sleep now. Lol

Answer 250

You can find the domain, but finding the range is always confusing for you

Answer 251

Your life would be a lot easier if you learned how to graph some of these

Answer 252

You need to learn how to graph the original function AND the inverse function.

Answer 253

Actually it’s the other way...cuz i know how to take the inverse of the original function and i can judge what x values will make the function undefined...so the domain of the inverse aka the range is easy for me it’s the freakin domain that eats my brains

Answer 254

Nah, the domain is easier than the range. I guess you won't know for sure until you take a test and get the results back.

Answer 255

You already know how to find a domain.

Answer 256

If an expression is in the denominator, then that expression cannot be zero. If an expression is under an even root then that expression must be greater than or equal to zero

Answer 257

It’s hard to find the domain of the original root functions for me... cuz i don’t know where to start from??? Do I plugin the range values of the original function back into the function to find the domain or what??

Answer 258

Easy, now to describe how to find the range (at least whenever it is not feasible) to use the inverse method, then describing the range might be more difficult

Answer 259

If you have an odd function \(f(x) = (x - 1)^{1/3}\) then the domain is all real numbers

Answer 260

I can find the range, help me find the domain of root functions... and weird functions..

Answer 261

If it's an even root \(f(x) = (x - 1)^{1/2}\) then set \(x - 1 \ge 0\) that's easy

Answer 262

Does having all real numbers mean i won’t have any restrictions in the domain ??? Is that always the case that i’ll Be all real numbers without restrictions ?

Answer 263

Trust me domain is easier than range

Answer 264

I find range easier than domain tbh ..

Answer 265

Okay let me prove you that the range is harder

Answer 266

Yeah, finding the domain of even root function is easy...but just gotta remember it applies to the inverse functions as well.

Answer 267

\(f(x) = \dfrac{\sqrt[4]{x - \sqrt{x}}}{(2x - 5)(3x + 8)}\)

Answer 268

Find the range of that

Answer 269

Bro, who even give questions like that for homework or tests?

Answer 270

I should have put a root in the denominator

Answer 271

Be a little more practical with your examples lol

Answer 272

You told me finding the range is easy. What's the problem now?

Answer 273

Bro, there’s a limit to *i find range easy* lol

Answer 274

right.

Answer 275

Finding the range ain't as easy as you think. That's the moral

Answer 276

Anyways...i gotta finish this up and sleep sleep... it’s almost 1 am here lol

Answer 277

\(\color{#0cbb34}{\text{Originally Posted by}}\) @TheConMan
Bro, there’s a limit to *i find range easy* lol
\(\color{#0cbb34}{\text{End of Quote}}\)
Sorry to interrupt..is the pun intended xD

Answer 278

Fully intended

Answer 279

Okay off to bed everyone

Answer 280

I think we've all overstayed our limit

Answer 281

K thanks for your help guys. I owe my math marks to wo guys

Answer 282

it's time to discontinue XD

Answer 283

^ this

Answer 284

We’ve reached our limits lol

Answer 285

Goodnight :)