Question

Help please

Answer 1

@dude

Answer 2

Oh, post the q

Answer 3

First one just asks about the description of the graph? I guess you can say that that its range cannot be negative (f(x))
For the second one the denominator cannot be 0, at least thats what I assume it means by that question

Do you know how to write the functions as, f(g(x)), g(f(x)) and so on?

Answer 4

Nope

Answer 5

for c would it be that the functions cannot be zero too

Answer 6

@Vocaloid

Answer 7

@Vocaloid

Answer 8

@dude

Answer 9

Im on can we continue please :)

Answer 10

f(g(x)) okay so the outside part is just the equation and whatever is inside you will substitute inside of the equation outside, for example the first one

\(f(x)=\sqrt{x^2-2}\) => substitute \(g(x)=\frac1x\) in for x

\(f(g(x))=\sqrt{{\frac{1}{x}}^2-2}\)

(Note: \(\frac{1}{x}^2\) is just \(\frac{1^2}{x^2}\) or \(\frac{1}{x^2}\))

so \(\sqrt{{\frac{1}{x^2}}-2}\)

Answer 11

Does that make sense?

Answer 12

Yes so for a-d we write the equation and descriptions?

Answer 13

It only seems to ask for the domain

Answer 14

ANd the domain would be the possible variables right?

Answer 15

All the possible x values, yes

Answer 16

okay so for a,

Answer 17

it would be : the domain of this function is all real numbers?

Answer 18

and its range cannot be negative?

Answer 19

\[(-\infty,0) U (0,\infty)\]

Answer 20

This is the domain actually

Answer 21

i mean i would this it is

Answer 22

? ya there

Answer 23

Not quite, the square root cannot be less 0 because it would be undefined
\(\sqrt{\frac{1}{x^2}-2}\ge 0\)
\((\sqrt{\frac{1}{x^2}-2})^2\ge 0^2\) which is just removing the root

\(\frac{1}{x^2}-2\ge0\)
Turn the -2 into a fraction with the same decimal by multiplying by \(x^2\) on both sides
\(\frac{1}{x^2}-\frac{2x^2}{x^2}\ge 0\)

\(\large\frac{-2x^2+1}{x^2}\ge 0\)

You can factor this to isolate the negative
\(\large \frac{-(\sqrt{2}x+1)+(\sqrt{2}x+1)}{x^2}\ge 0\)
Multiply both sides by -1 (inverse the inequality)
\(\frac{(\sqrt{2}x+1)+(\sqrt{2}x+1)}{x^2}\le 0\)
Now just redistribute the factors
\(\frac{2x^2-1}{x^2}\le 0\)
You now just want to make a (sign diagram) find when \(2x^2-1\) negative as well as \(x^2\)

\(2x^2-1=0\)
=>\(2x^2=1\)
=>\(x^2=\frac12\)
=>\(x=\pm\sqrt{\frac12}\)

and

\(x^2=0\)
=>\(x=0\)

Anyway, made this too long but basically
\(\mathbb D:[-\sqrt{\frac{1}{2}},0) ~U~(0,\sqrt{\frac{1}{2}}]\)

Answer 24

okay so I think I understand this. Would my statement about the range not being negative be valid

Answer 25

Yes its valid

Answer 26

okay for the next one the domain would be the opposite then

Answer 27

\(g(f(x))=\frac{1}{\sqrt{x^2-2}}\)
From this we know that the denominator has to be greater than 0
\(\sqrt{x^2-2}\ge0\)
\(x^2-2\ge0\)
\(x^2\ge2\)
\(\pm\sqrt2\)

\(\mathbb D:-\sqrt{2}>x ~and~x>\sqrt{2}\)

Answer 28

question

Answer 29

does the zero cancel out?

Answer 30

Where

Answer 31

when you get from x^2-2-0 to x^2>2

Answer 32

\(x^2-2\ge0\)
\(x^2\ge2\)
\(\pm\sqrt2\)

this part?

Answer 33

oh bcuz zero doesnt satisfy the function?

Answer 34

yes

Answer 35

\(\sqrt{x^2-2}\ge0\)
So I first raised both sides to the second power so the root cancels and I have to do 0^2
0^2 is just 0, it doesn't cancel itself
\(x^2-2\ge0\)
\(x^2\ge2\)
\(\pm\sqrt2\)

Does that answer your question?

Answer 36

Yes it does .

Answer 37

forc we wouldnt use 1/x in our function

Answer 38

so would the f equation just double like be used twice

Answer 39

Right

Answer 40

\(f(f(x))=\sqrt{\sqrt{x^2-2}-2}\)

Answer 41

Same with the fourth one
\(g(g(x))=\frac{1}{\frac{1}{x}}\) which is just x

Answer 42

okay give me a sec while I figure out the domains

Answer 43

Okay

Answer 44

\[(-\infty,-\sqrt{6}] U [\sqrt{6},\infty)\]

Answer 45

Yes

Answer 46

Yay, and for an explicit description would it be that no negative numbers?

Answer 47

and then for d it would be \[(-\infty,\infty)\]

Answer 48

so all real numbers

Answer 49

Yes to both

Answer 50

Yay

Answer 51

woot

Answer 52

For rational functions, the denominator is ≠0

Answer 53

for this one,,,

domain for p(x) \[(-\infty,1)U(1,\infty)\]

Answer 54

Right

Answer 55

domain for q(x) = \[(-\infty,\infty)\]

Answer 56

Yeah, no restrictions there

Answer 57

okay for b how would I show that

Answer 58

Set them equal to each other
\(\large \frac{x^3-x}{x-1}=x^2+x\)

Answer 59

okay and solve?

Answer 60

Yes

Answer 61

I got all real numbers

Answer 62

for do I plug in 1?

Answer 63

Yes

Answer 64

\[\frac{ 1^3-1 }{ 1-1 }1^2+1 = false\]

Answer 65

Well the issue is when you substitute 1 in the denominator, remember we wrote the range
\(\large \frac{x^3-x}{x-1}\) x could not be 1 because it makes the denominator 0

So trying to substitute would automatically make it undefined
\(\large \frac{1^3-1}{\color{red}{1-1}}=1^2+1\)

Answer 66

okay and for c : yes, the functions are the same because when you combine them you end up with the same answer

Answer 67

Right

Answer 68

okay: and I already sketched it so next question

Answer 69

x=1+e?

Answer 70

Okay, have you tried solving or are you lost?
Also I havent done logs in some time, I'm rusty

Answer 71

I solved and got x=1+e

Answer 72

For the first or second equation? o.o

Answer 73

1st

Answer 74

I got something different, what did you do?

Answer 75

i simplified , and then isolated

Answer 76

its all written out and its hard cause i cant take a pic

Answer 77

Ah its okay

Answer 78

SO are we good with my answe?

Answer 79

@Angle

Answer 80

x=1+e is correct

Answer 81

Yay

Answer 82

and for the next equation x=1+sqrt2

Answer 83

I believe there should be two answers

Answer 84

for the first equation

Answer 85

?

Answer 86

for the second equation
x=1+sqrt2 is correct
but I think there is another one

Answer 87

1+(-sqrt2)?

Answer 88

yup

you can sum it up into one answer with x = 1 ± sqrt(2)

Answer 89

perfect. and just need help with one more

Answer 90

if you can

Answer 91

sure

Answer 92

here's an example

Answer 93

so 0.002/1/2

Answer 94

mmm it's called "half" life, but it doesn't have to be 'half'

Answer 95

oh okay so the half life is 500

Answer 96

Thank you angle !!!!

Answer 97

no problem

haha hopefully I'm not wrong ^_^"