Question

let f(x)=sqrt 7x and g(x)= x+8
what is the smallest number that is in the domain of (f o g)

Answer 1

@pooja195 do you think you could help me?

Answer 2

@DanJS

Answer 3

This is similar to your previous question, please set it up as f(g(x))

Answer 4

Okay

Answer 5

@Nnesha

Answer 6

would that be sqrt(x+8)?

Answer 7

\[\huge f(g(x)) = \sqrt{7(x+8)}\]

Answer 8

Now how do you find the domain of this composition?

Answer 9

Plug in numbers?

Answer 10

Well, we can't have a negative number in the square root right?

Answer 11

No it can't be negative

Answer 12

Good, so simply we take what's inside our square root \[\huge 7(x+8) \ge 0\] solve for x :)

Answer 13

but could my x for now be a negative to get say sqrt 7(-6+8) something like that and have 7(2) sqrt

Answer 14

or would it not work

Answer 15

That does work, but plugging numbers would take forever, and if you solve for x, we can write our domain as it gives us a clear picture of what x's are allowed and what aren't. So go ahead and solve for x and you will see yourself!

Answer 16

I have to get the smallest domain I can remember

Answer 17

That's ok, this will give you your domain. And we'll check after by plugging in numbers

Answer 18

To convince you

Answer 19

Negative four would be the smallest tho

Answer 20

Please, solve for x

Answer 21

How would you suggest I do that?

Answer 22

\[\huge 7(x+8) \ge 0 \] lets ignore the inequality sign for now, how would you solve for x if I gave you \[7(x+8)=0\]

Answer 23

-8 right because

Answer 24

\[7(-8+8)=0 7(0)=0\]

Answer 25

Exactly! So we have with our inequality \[x \ge - 8\]

Answer 26

Okay now we have a starting point correct?

Answer 27

Well I would have hoped you do it algebraically, as that's what important, so to do it that way \[\frac{ 7(x+8)}{ 7 } \ge \frac{ 0 }{ 7 } \implies x+8 \ge 0 \implies x +8-8 \ge 0-8 \]\[\implies x \ge -8\]

Answer 28

Good?

Answer 29

okay slow down slightly

Answer 30

Ok no worries, did you get lost anywhere during the algebraic process?

Answer 31

Yes sir

Answer 32

Alright so our goal here was to solve for x (get x on its own) ok?

We started off by dividing both sides by 7 as you can see the 7 is "attached" to the (x+8) meaning it is being multiplied, so we use the opposite operation which is division to remove it as it will turn it into 1 on the left side and 0/7 of course is 0.

Answer 33

okay I get that part

Answer 34

Ok the next part we had \[x+8 \ge 0\] similarly notice how we are adding here on the left side, so again we use the opposite operation of what already exists and apply it to both sides. So we subtract -8 on both sides \[x+8 \color{red}{-8} \ge 0 \color{red}{-8} \implies x \ge \color{red}{-8}\] good?

Answer 35

Yeah thanks for changing colors to show me btw

Answer 36

No worries, now let me ask you a question.
What value is bigger -8 or -4?

Answer 37

-4 is bigger

Answer 38

Because its closer to zero

Answer 39

Right! :)

Ok notice since we solved for x there, that inequality means x must be `greater than or equal` to -8, so that will be our smallest value allowed.

Answer 40

okay but can the domain of f o g be a negative?

Answer 41

because -8+8=0
Then 7(0)=0
sqrt(0)=0
would that work?

Answer 42

Yes, that is what we were solving for, don't confuse it with your last question because we had \[x-1 \ge 0 \implies x \ge 1\] in your last question

Answer 43

^ that's from your last question

Answer 44

If I can recall it correctly haha

Answer 45

That is but he told me it has to be greater then 1

Answer 46

Ah gotcha, it was a denominator as well yes then it would be greater 1!
Also, you are allowed to have 0 under your square root

Answer 47

Okay thank you for clearing that up

Answer 48

Yw :)

Answer 49

So my smallest number would be -8 correct because the square of 0 is 0

Answer 50

Correct, essentially when you have a square root, we take what's inside it and set it up an inequality as we did before and that will give you your restrictions for your domain.

Answer 51

okay thanks

Answer 52

:)

Answer 53

The answer was correct thanks a lot :D

Answer 54

now I have a tough question lol