Question

I have a college algebra chapter test tomorrow and I'm reviewing through some problems in my textbook. Two of the problems are confusing and I was hoping someone can go through it with me.
Let f(x)=square root of x, g(x)=x-4, and h(x)=1 over x-2. Find an equation defining each function and state domain of the function.
I'm doing problem 29, which is f-h. Also problem 33, which is g/f.

Answer 1

what have you determined so far?

Answer 2

29.) as far as I got is \[f-h=\sqrt{x}-\frac{ 1 }{ x-2 }\]

Answer 3

seems fine to me, what do you know about domains and ranges

Answer 4

I know domains are the x's and the ranges are the y's. But I'm not sure how to get the domain from this problem

Answer 5

lets refine that:

domain are all the usable xs. all x values that do not make any bad maths

the range is all the ys that we can get from the domain vlaues

Answer 6

can you think of anything in your f-h that would make the math go bad?

Answer 7

think of things that make us divide by zero, and make us take the sqrt of a negative number

Answer 8

I don't understand what you're asking me to do.

Answer 9

look at the equation you made:\[f-h=\sqrt{x}-\frac{ 1 }{ x-2 }\]
the domain is the set of all z values that make good math sense.

you have heard thru the years that there are things that math cant do, we can divide by zero is one of the cardinal rules of math. another deals with sqrt functions and log functions. those are the big ones that come to mind

Answer 10

*all x values ... my fingers hate me

Answer 11

So would it be 2?
Sorry, I lost connection and the screen froze.

Answer 12

notifications seem to be on the fritz as well.

Answer 13

x=2 is a bad value, and anything less than 0

Answer 14

so we define the domain as: [0,2) u (2,inf)

Answer 15

So basically, you had to find a value that's bad?

Answer 16

basically yes, we had to determine what values of x we simply cant use

Answer 17

Ohhh okay

Answer 18

Thank you so much. So quick question, the infinity sign is to define what we can use?

Answer 19

infinity is an unbounded direction. when we use infinity in our set, it means that there is nothing else between here and forever that bothers us

Answer 20

Oh okay, so for problem 33, which is g/f: The answer is \[[0, \infty)\] because you can't use 0 and anything pass 0?

Answer 21

Let f(x)=square root of x, g(x)=x-4,

x-4
------
sqrt(x)

your almost correct, the notation is a bit off but i have a memory device for that

Answer 22

if something is equal to, then make an equal sign



and connect it by a bar to make it a bracket

Answer 23

if its not equal, then its not this bracket

Answer 24

Ohhh, so since it's not equal to, the answer is \[(0, \infty)\]

Answer 25

correct

Answer 26

Wouldn't 1 be a good value for g/f becauase -3 over 1 is -3?

Answer 27

1 is fine, the notation tells us what we can use

use all x values that are between 0 and inf. 1 is a value between 0 and inf

Answer 28

.00000001 is also fine to use

Answer 29

Okay so let's go back to problem 29, which is f-h. So the answer is \[[0, 2) \cup (2, \infty) \] because why?

Answer 30

sqrt(x) is bad for x < 0; its domain is [0,inf)

1/(x-2) is bad for x=2 so we have to remove it from the domain giving us

[0,2) 'unioned with' (2,inf) 2 is not useable so it doesnt have the equal brackets

Answer 31

1.9999999999999 is usable

2.00000000000001 is usable

its just simplest to split the set into two parts, to the left of 2, and to the right of 2

Answer 32

Okay i understand, thank you so much Amistre64 :)

Answer 33

your welcome