Question

What is the domain of the compound funciton f(x)/g(x) if x is any real number, g(x) = x -1 and f(x) = f(x) = (x+1)1/2
walk through, please?

Answer 1

@Directrix if you get a second, you're usually really helpful!

Answer 2

I don't follow this: >> f(x) = f(x) = (x+1)1/2

Why two f(x) terms there?

Is (x+1)1/2 supposed to be (x+1)/2 or something else.

Answer 3

No idea why f(x) is written twice... it might be a typo. I'll send a picture

Answer 4

That's the whole explanation. I understand why x cannot equal 1, but not sure about the x>= -1?

Answer 5

Anyone?

Answer 6

Look at your compound function f(x) / g(x).....g (x) is the denominator correct?

Answer 7

Not sure what you mean...

Answer 8

You see in the question you have a compound function f(x) / g(x)..?

Answer 9

Like when you just put the functions in, you write is as ((x+1)1/2)/ (x-1), right? So then x cannot equal 1

Answer 10

Correct because 1-1 would be 0 and you cannot have 0 as a denominator.

Answer 11

Okay, but the answer is x >= -1 and also x does not equal 1

Answer 12

so I am confused about the x >= -1?

Answer 13

@freckles if you have a second, any idea about this one?

Answer 14

first let me ask do you know how to evaluate:
\[(-5)^\frac{1}{2}?\]

Answer 15

can you do the principal square root of a negative number?

Answer 16

I don't really remember to be honest. I can look it up to refresh my memory.

Answer 17

principal square root of negative numbers don't exist over real numbers

Answer 18

\[\sqrt{x} \text{ has domain } x \ge 0\]

Answer 19

\[\sqrt{x+1} \text{ has domain } x+1 \ge 0\]

Answer 20

we want the inside to be positive or zero

Answer 21

that is what the greater than or equal to zero means

Answer 22

\[f(x)=\sqrt{x+1} \text{ has domain } x+1 \ge 0 \implies x \ge -1 \\ \text{ so domain of } h(x)=\frac{\sqrt{x+1}}{x-1} \text{ is } x \ge -1 \text{ and } x \neq 1 \\ \text{ the } x \neq 1 \text{ comes from the bottom can't be zero }\]