# Domain and range question

1. Apr 2, 2005

### joejo

domain and range question!!

Hi,

I have a quick domain and range question, I have placed my answer below the question. I'm having doubts that it is right! Can someone please help me verify it. Thanks in advance!
_____
Let f(x)= x² + 2 and g(x) =√1-x²

a) Find the domain and range of f and g.
b) Are the functions f.g and g.f defined? Explain?

The domain of f is {X|X ε R}
The range of f is {y|y ≥ 2, yε R}

f.g= f[g(x)]
= f ___
(√ 1-x²)
= ___
(√ 1-x²) + 2
= 1- x² + 2
= 3-x²

All of the above is correct. I have also already done g.f as well.

This is where i need help....what comes next....
Are the functions f.g and g.f defined? Explain? and what is the domain of g and the range of g?

Last edited: Apr 2, 2005
2. Apr 2, 2005

### whozum

They are defined, because they both exist.

The domain of g is all numbers less than and including 1, because once you get alrger than one, you will be taking the square root of negatives.

3. Apr 2, 2005

### joejo

hey thanks for the quick reply, but how about the range of g?

4. Apr 2, 2005

### whozum

I gave you a huge hint on the lower limit of the range of g, and knowing the domain of g you can figure out the lupper limit on the range of g.

5. Apr 2, 2005

### joejo

well this is what i have so far,

the doman of g is {x||x| ≤ + 1}
the range of g is {y|y ≥ 0 ≤ y ≤ 1}

g of f isn't defined because domain is not real?!!

is that right

6. Apr 2, 2005

### whozum

It doesnt need to be absolute x if your squaring it.

Correct
$$g=\sqrt{1-x^2}$$

$$f=x^2+2$$

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

So (x^2+2)^2 must be less than 1, work from here.

7. Apr 2, 2005

### joejo

sorry you're losing me...ive been trying for the past 15 minutes....anyone else or whozum can you please clairfy

Last edited: Apr 2, 2005
8. Apr 2, 2005

### Data

Actually it does... precisely because you are squaring it! Look again :tongue:

And while "range of $g = \{ y | y \geq 0 \leq y \leq 1 \}$" is correct, it's also a pretty strange way of writing it. You can just write $\{ y | 0 \leq y \leq 1\}$.

9. Apr 2, 2005

### whozum

What are you missing

10. Apr 2, 2005

### Data

You are quite right that $g \circ f$ isn't defined over $\mathbb{R}$ for any real $x$.

Whozum's anaylsis was right, but then again so was your comment he responded to: $(x^2 + 2)^2 \leq 1$ is not satisfied for any real $x$.

11. Apr 2, 2005

### joejo

so is this right?

the domain of g is {x||x| ≤ + 1}

12. Apr 2, 2005

### Data

indeed it is.

13. Apr 2, 2005

### joejo

so it's not.......

the domain of g is {x| x ≤ + 1}

14. Apr 2, 2005

### Data

well, is $g(-2)$ defined? Is $-2 \leq 1$? That should answer your question! :tongue:

15. Apr 2, 2005

### whozum

x^2 is always positive. saying |x|^2 is redundant, isnt it.

16. Apr 2, 2005

### joejo

no its not!

17. Apr 2, 2005

### joejo

data no it isn't....

18. Apr 2, 2005

### whozum

Oh ok. I see what you mean by absolute. I was thinking x can go from -1,1 which is the same as |x| from 0 to 1

19. Apr 2, 2005

### Data

Yes it is. But precisely because of that,

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

is real if and only if $|x| < 1$. If $|x| > 1$ then either $x < -1$ or $x > 1$. In either case, $(-x)^2 = x^2 > 1$ so $1 < x^2 \Longrightarrow 1 - x^2 < 0$.

20. Apr 2, 2005

### whozum

how come your latex looks so sexy