Question

can someone algebraically show me how to plug and chug for every x value into the function? i'm having hard time dealing with power functions. wait let me draw tho.

Answer 1

If you want to find the value of f(x) when x is, say -3, here's what you would do:

f(x)=3x- 1/3
You want to find when x=-3. The algebraic notation for this is f(-3).
Replace the x variable with the value, -3.

3(-3) - 1/3
You can try this and you'll get your answer. Apply the same thing for the other values.

Answer 2

what do u mean *apply the same thing on the other side*? be more specific plz :)

Answer 3

"Apply the same thing for the other values"
Do the same for
f(-2), f(-1), etc.

Answer 4

im having trouble with powers... ..wait i'm drawing it out for u. 1 sec

Answer 5

Oh, I couldn't tell that was an exponent.

Answer 6

That changes all the values. Ignore what I said earlier


Let me explain:

Answer 7

For the sake of Santa clause...i’ll Draw out the question more clearly. 1 sec

Answer 8

No, you made a mistake in the second step. The power applies to the entire binomial.

Answer 9

Ok wait, let me try one more time... wait and check my second attempt

Answer 10

Recall that if a power is a fraction, the numerator is the power and the denominator is the index of the radical.

Answer 11

Sure

Answer 12

Don’t do the question for me. Let me struggle myself. Just correct when when am wrong

Answer 13

Sure, do you want a suggestion though?

Answer 14

You're getting a step closer. Now the '4' should go inside the radical and around the binomial.

Answer 15

First comes power, then comes root.

Answer 16

What do u mean by *the denominator is the index of the radical*?

Answer 17

So do I switch the 4 and 7?

Answer 18

Index of a radical is like

\[\sqrt[index]{n}\]

Answer 19

What you're rooting it by. In this case it's 7

Answer 20

Oki dokie

Answer 21

Wait, let me write it out again

Answer 22

Sure

Answer 23

Correct.

Answer 24

Omg I’m shoooo happy. I’m finally learning to do these without the calculator lol thanks mate...wait let’s continue 😀

Answer 25

np

Answer 26

Do I take the sq root of q23 and then multiply by 7?

Answer 27

Do you have an option in your calculator?

Answer 28

No you wouldn't do that

Answer 29

One second. Let me try that. U confirm if it’s right or wrong. I jus wanna see what happens

Answer 30

Sure

Answer 31

77.7.6 lol

Answer 32

What single steps do I plugin into my calculator to find the right answer?

Answer 33

Okay that would be wrong

Answer 34

let me check your calculator model and see

Answer 35

Is it a scientific/graphic calculator or normal?

Answer 36

Ofc it’s a scitific calculator mate 🙂😇

Answer 37

But jus really really old

Answer 38

There should be an option like
\[\sqrt[x]{ }\]

Answer 39

I have an option of....wait let me draw

Answer 40

Maybe look up a tutorial for it and figure out where that option is. I don't have a casio calculator so I'm not sure about how they work

Answer 41

sure

Answer 42

Wait. What’s the answer mate?

Answer 43

It's okay, I think you can still do it without a calculator.

Answer 44

The answer is like 3.5

Answer 45

3.58

Answer 46

Whole dawn... let me try some stuff

Answer 47

Here's what you'd need to do without a calculator:


Apply a principle of exponents:

\[(\frac{ x }{ y })^{n}\]
=
\[\frac{ x^n }{ y^n }\]

Answer 48

Sure

Answer 49

So you would do that for the radicand.

Answer 50

No, let's go back to where we had

(-10/3)^4

Answer 51

\[\sqrt[7]{(\frac{ -10 }{ 3 })^4}\]

Answer 52

Apply the principle I just showed you.

Answer 53

No, only for the radicand. Don't do the 4/7

Answer 54

It should be

\[(\frac{ -10 }{ 4 })^4\]

Answer 55

We are applying this principle within the radical

Answer 56

It's fine

Answer 57

Ok ok ok ok... so right right we r only solving what’s within the root?

Answer 58

For now, yes.

...Also, I just noticed you made a mistake WAY earlier

Answer 59

Wait, let me take the first stab at this plz

Answer 60

Yes yes but wait a second. You got the radicand portion wrong and I failed to notice. My apologies.

-9-1/3 isn't -10/3

Answer 61

It's -28/3. You didn't take the correct common denominator.

Answer 62

-9 would be -27/3.

Do not fret, these mistakes occur often. Just make sure you don't take simpler portions of a problem lightly.

Answer 63

So now you have

(-28/3)^4

Answer 64

\[\sqrt[7]{(\frac{ -28 }{ 3 })^4}\]

Without using a calculator, we would first evaluate the radicand.

Answer 65

Correct, you're on the right path. Keep going

Answer 66

But the parenthesis should go to the denominator as well

Answer 67

Correct

Answer 68

Now apply the principle I showed you earlier to the radicand

Answer 69

Correct

Answer 70

Correct

Answer 71

Yes. That is also correct, you're just switching the form.

Answer 72

Got em!

Answer 73

Do you want to simplify it more?

Answer 74

Now here’s the issue with my calculator....watch.

Answer 75

It won't be much, and usually, this form is accepted as well.

Answer 76

Sure

Answer 77

I need to right down this whole question on a paper. Brb. Your a legend mate. I think I should call u Jesus from here nwo

Answer 78

On*

Answer 79

I have more fraction power questions tho. I was suppose to find their domains and ranges ....i was plugin in different ‘x’ values into the function to plot the graph...and my crappy calculator would gimme stupid answers. Thanks tho

Answer 80

Ah okay. For domain and range I suggest you take the limits (RHL and LHL) of them or it would take too long. There's 1 discontinuity there.

Answer 81

Now I’m getting the right answers. Wait let me post the picture lol

Answer 82

Or you should graph them. Using test points here would take too long as the discontinuity is really small and hard to pinpoint with a test point.

Answer 83

sure

Answer 84

before, everytime i plugged in *-5^2/3* i'd get an error lol

Answer 85

Calculations are correct, but using test points for these types of complex functions would take too long just saying. Also if there's a discontinuity, you won't be able to pinpoint it with test points.

Answer 86

do i need to buy a better calculator or something?? to be able to get the *y* values faster?

Answer 87

Do you have a graphing calculator? It would help you tons to find the domain and range. Otherwise, you'd need to use calculus.

Answer 88

i dont wanna buy a fancy TI tho. cuz the uni wont let me use it in midterm and finals...sooo

Answer 89

Do they allow a TI-83/84

Answer 90

It's a pretty basic graphing calculator

Answer 91

pff, they won't even let me enter college with that calculator mate

Answer 92

TI-83 or TI-84??? u can literally graph anything on those things

Answer 93

weight, i have more questions. i'll be back in 2 minutes. gotta finish this question :)

Answer 94

sure

Answer 95

this is the graph of that function. https://www.desmos.com/calculator/xno3esxjcq

Answer 96

can u see it?

Answer 97

So the domain is all real numbers?

Answer 98

Yes.

Answer 99

What about range? All real numbers except y<0

Answer 100

Sorry y has to be greater than 0

Answer 101

Correct but y≥0

Answer 102

0 is inclusive

Answer 103

Yes my bad. I’m doing it with u on the fly

Answer 104

Just saying, using test points won't always work. For example, if you get a question like:

\[\sqrt{x^2 + 1/3}\]

There would be a discontinuity

Answer 105

sure

Answer 106

I actually do have questions like that and even more funnier looking stuff

Answer 107

okay, yes here you would then need to know asymptotes and discontinuities of functions.

Answer 108

Testpoints don't always work.

Answer 109

Because there might be a discontinuity all the way at x=928, for example

Answer 110

So what variables in the numerator and denominator decide where the asymptote(s) and discontinuities r

Answer 111

So for this question...

Answer 112

I'll explain:

Answer 113

For that function, there is no discontinuity

Answer 114

I know I can’t plugin 2 into the equation...cuz that’d make everything equal to zero

Answer 115

Just because the output is 0 doesn't mean it's a discontinuity.

Answer 116

So for that reason, we have a restriction in the range? Where y can’t be bigger or equal to zero.

Answer 117

Yes, I should not have used that question for discont and asymp

Answer 118

For this question, you were correct. I'm just saying for hypothetical scenarios if you get a question that has a discontinuity, test points don't work. There's a w

Answer 119

There's a way* to calculate asymptotes and discontinuities. Do you know them?

Answer 120

i think I do but wait...I have a question. When we plugging zero into the function..the whole function collapses. How do I know it’s for the domain or range....if I were to do it really fast

Answer 121

I do have asymptotic and restriction questions ...but i’ll Ask u those later..

Answer 122

I can’t make out what u just sent me...it all blurry

Answer 123

For domain it's the set of x values. As long as you yield an output that is not n/0, the input exists.

For range, it's simply knowing the minimum or maximum and going to their approaches

Answer 124

Anything n/0 is undefined..got it

Answer 125

Yes. Otherwise, the input is considered in the domain.

Answer 126

Also, if the radicand is less than 0, it is also imaginary and thus undefined.

Answer 127

ie
\[\sqrt{x-4}\]

If you get a value under 4, you'll get a negative radicand.
That's why the domain is x≥4.
Range would be y≥0. You can use some testpoints here. Also, it would be best to know the behavior of these functions. Square root functions normally have a range greater than or equal to 0 unless it is transformed by f(x+h)+k

Answer 128

So if a function has a y value for any x value, it’s automatically in the horizontal domain? So if there’s no value for x...let’s say at f(2) of some function gives me an error ....means there’s an asymptote

Answer 129

Yes I agree with the root question...where x has to be bigger than 4 bcuz u can’t take the root of negative numbers

Answer 130

So does that mean I need to remember than root function graphs look like? Omg ok

Answer 131

Discontinuity, not asymptote. There's a thin line between the two. Discontinuity is a specific point, asymptote is more like a region or line.

Answer 132

And some discontinuities are removable, which is why you should know to manipulate functions

Answer 133

Yes. Like in limits...I got it

Answer 134

It's best to know the basic look of them so you can figure out if the function is transformed. Then you wouldn't need to do any work.

Answer 135

And no, not for all root functions, I'm just saying for like square roots. I myself don't know the behavior for cube root functions and others

Answer 136

Yes, in limits

Answer 137

I’m working on a new question. I know wanted to see what it’s like to manually solve for the y value for this function. Picture coming soon

Answer 138

Bad idea to use test points here. Just look at the denominator and you can get a discontinuity.

Answer 139

asymptote*

Answer 140

i just wanted to test my algebra tho :)

Answer 141

oh ok

Answer 142

i think the denominator has to do with vertical asymptote? O-o

Answer 143

Correct but then there's a vertical stretch in the numerator

Answer 144

hoesntly i havent touched math in years... im just guessing/recalling to the best of my abilities.. plz dont judge..

Answer 145

k so before we figure out the domain and range of the following function. And the vertical and horizontal asymptotes ...can u plz check my work real quiff? let me draw :)

Answer 146

sqrt(-4) is imaginary and thus undefined, you can't factor out a -2.

Answer 147

Oki 😐

Answer 148

That's why x=-3 for f(x) is undefined.

Answer 149

You can't solve any further

Answer 150

oki dokie

Answer 151

so the vertical asypmtote is at 1?

Answer 152

since 1 will make the denominator equal to 0

Answer 153

correct

Answer 154

any anything divided by 0 is error

Answer 155

That makes the domain (1, infinity)

Answer 156

and for the horizontal asymptote... we look at the ratios of the leading degrees. wait, let me draw

Answer 157

yes correct

Answer 158

wait.... i need to show u the graph... :O

Answer 159

but it would be an oblique asymptote

Answer 160

watinda heck is this?

Answer 161

the horizontal asymptotoe is more like at 1 tho

Answer 162

No, if the degree in numerator is > than denominator's, it's oblique/slant asymptote. It would be difficult to calculate here.

Answer 163

but i just watched the video on youtbe and he said it's *0* tho *~*

Answer 164

https://www.youtube.com/watch?v=WKwc0sxLRTs stop at 2:54

Answer 165

omg i mixed up... it's the opposite here. we have a bigger degree in the numerator than the denominator. sorry *~*

Answer 166

omg he said, if the degree of the numerator is bigger than the denominator, then there's no HA

Answer 167

at 5:35

Answer 168

yes it would be oblique, but in this case it can't be calculated as all the possible lines would be tangent

Answer 169

so the horizontal and oblique asymptote is null

Answer 170

so what do we write? 0?

Answer 171

you don't need to write or just say null

Answer 172

oki dokie :)

Answer 173

k the domain is all real numbers, but x has to be bigger or equal to 1

Answer 174

yes

Answer 175

as for the range, it'd be all real numbers except, y can't be equal to or less the 4

Answer 176

okie dokie :)

Answer 177

yes

Answer 178

It can be 4, actually.

Answer 179

wait, what do u get when you plug in x = 2

Answer 180

let's do it here actually. let me draw

Answer 181

Ok, it can be 4

Answer 182

D:
\[X \in \mathbb{R}, x>1 \]
R:
\[y \in \mathbb{R}, y≥4\]

Answer 183

yes

Answer 184

I think I helped enough, someone else can take over or just post another q as I have to go.

Answer 185

thanks bye

Answer 186

this is a rational function and i need to find the domain and range of the function.

Answer 187

I’m trying to factor this bad boy for the past 3 minutes...let me post

Answer 188

So you can't factor the denominator because it is not following the rule (a-b)^2 unless you use complex numbers. Do you know how to do that?

Besides, you wouldn't need to factor the denominator.

Answer 189

You can't factor it using real numbers/variables.

Answer 190

Okie dokie

Answer 191

If you factor it, the expression would look something like: (x-i)(x+i) where i=sqrt(-1)

Answer 192

Soooo how do we go about finding the vertical asymptote?

Answer 193

There is no vertical asymptote.

Answer 194

Yuhh, that’s what I get ;/

Answer 195

I mean there is, but it is imaginary and we cannot graph imaginary scalar or vector quantities on the cartesian plane.

Answer 196

But how do I make myself sure that there isn’t one?

Answer 197

And asymptotes only fall under the real number category, so we might as well say that this function has no vertical asymptote.

Answer 198

So

If a variable in the denominator has the ability to make the denominator 0 at any given point, there is a vertical asymptote.
If the variable in the denominator has imaginary/complex solutions, there is no vertical asymptote.

Answer 199

No.

Let's say a denominator for any given rational equation/expression is
x^2 +4.

Can you tell me the roots for x?

Answer 200

So when I can’t factor the deno, do I just assume there r no asymptotes?

Answer 201

Basically for vertical asymptotes:

1. Look at the denominator only. Set the polynomial in the denominator to 0. Solve for the variable as you normally would using algebra.
2. If you get real solutions, there is a vertical asymptote and it would be that solution.
3. If you get imaginary solutions, there is no vertical asymptote.

Answer 202

I can try to factor that...weight, let me draw it out..one sec

Answer 203

No, just because you cannot factor it does not mean there is no solution. It just means the polynomial is 'prime' or cannot be factored.

The decision between no asymptotes and asymptotes is decided if there is a real or imaginary number.

Answer 204

Okie dokie. I’ll note that down. Good point tho.

Answer 205

Correct but you don't need to factor. You can if you want. But what I would look for is a way to see how to isolate the variable.
When you do that, there would be a negative number over a square root, which cannot be graphed.

Answer 206

Yes yes yes that’s I left it...

Answer 207

Brb, let me include that in my notes

Answer 208

Sure

Answer 209

As for the horizontal asymptote....we know that the leading highest power in the numerator is much smaller than the leading power in the denominator...thus...the horizontal asymptote is y=0

Answer 210

I'd just like to add on a point, though. In some cases, you can use manipulation and remove the imaginary component to the denominator and find an asymptote/discontinuity. Similar to limits.

Answer 211

I don’t need to lern that ....

Answer 212

if numerator<denominator then y=0
numerator > denominator no HA. but there is an oblique
numerator = Denominator then HA = ratio of leading coefficients.

Answer 213

correct

Answer 214

Let me actually include that in my notes too...brb

Answer 215

Back...so the domain is all real numbers

Answer 216

In this case? Yes

Answer 217

And since there’s no V.A., ........did u see the graph tho?

Answer 218

It doesn’t go above 2 I believe

Answer 219

Wait.. le me draw

Answer 220

So how do we figure out the point of vertex?

Answer 221

Sorry I have to go i can help later.

Answer 222

Okie dokie. Have a good one