Question

how do you fine the domain of a function?

Answer 1

The domain of a function f is the set of all values 'x' such that f(x) is defined.

Answer 2

So it is basically the list of 'legal' inputs for the function.

Answer 3

\[f(x)= 10x-4 \over x-1\]

Answer 4

i am confused on how to solve it and isnt really clicking in my head

Answer 5

You will need to set the denominator equal to 0 and solve because the denominator cannot equal 0
x-1=0
x=1
so the domain is all real numbers except one

Answer 6

You know that you are not allowed to divide by 0 right? So any value of x that makes a 0 in the denominator would be not allowed and therefore not in the domain.

Answer 7

well if x is 1 that wont it make 10-4 equal 6 so will that still work out?
so in order to make this questoin correct the answer would have to be all real numbers greater than 1

Answer 8

If x =1 you have:
\[\frac{10-4}{1-1} = \frac{6}{0} = \text{Undefined (cannot divide by 0)}\]

Answer 9

\[6 \over 0 \] is fine the way it is? and what makes you solve the bottom x then the top x

Answer 10

You have to do them both.

Answer 11

You cannot have a 0 in the denominator and the rule for finding the domain of in x with a fraction is by setting the denominator=0

Answer 12

but isnt the answer all real numbers greater than 1?

Answer 13

No because any number other 1 works

Answer 14

can you be clear on your response?

Answer 15

When i say works, I mean the denominator does not equal 0. But when you plug in 1, the denominator equals 0, but any other number does not bring it to 0

Answer 16

so in order to solve the problem i listed, i have to make the denominator a 0 to work and the ONLY number that can do that is 1 so therefore the answer or domain can only be 1. is that correct?

Answer 17

The denominator cannot equal 0, that is a no no. The domain is all real numbers except for 1. When you are trying to find the domain of a fraction you set the denominator equal to 0 and solve similar to the example above.

Answer 18

The domain would be all numbers that do NOT make the denominator equal 0.

\[\implies x -1 \ne 0\]\[\implies x \ne 1\]

Answer 19

so it has to be all real number or greater numbers than 1? in order to find my equation

Answer 20

Almost, the answer is all real numbers except 1, this includes negatives

Answer 21

It has to be all real numbers that are not 1.

\[x \in \mathbb{R} \setminus \{1\}\]

Answer 22

ohhh ok i understand now well at least this problem i do because know i know for sure that 0 CAN NOT be a denominator

Answer 23

Or if you want interval notation:
\[x\in (-\infty,1) \cup (1,\infty)\]

Answer 24

Haha that is good, as long as you know 0 cannot be in teh denominator you should be good

Answer 25

ok then let me toss you this problem because in know that type of question for sure

Answer 26

Also you can't take the square root of a negative. That's another one they like to use for this.

Answer 27

then how would you solve \[ f(x) =\sqrt{-x}\]

Answer 28

?

Answer 29

So the part under the radical cannot be negative:
\[\implies -x \ge 0\]\[\implies x \le 0\]

Answer 30

if it cant be negative then how do i change it so it can be possitive?

Answer 31

So x must be a negative number or 0 in order to give a valid square root.
e.g.

\[x = -9 \implies f(x) = \sqrt{-(-9)} = \sqrt{9} = 3\]
But
\[x = 25 \implies f(x) = \sqrt{-(25)} = \sqrt{-25} = \text{Not allowed.}\]

Answer 32

ok so all numbers have to be negative and how would i phrase it to say that the domain is all the negative numbers?

Answer 33

?

Answer 34

You can use interval notation..
\[x \in (-\infty,0]\]

Answer 35

how about words? i want to get in a habit in reconginizing of what to write comes to SAT