# Domain and range question

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:

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
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.

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

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.

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

joejo said:
well this is what i have so far,

the doman of g is {x||x| ≤ + 1}
It doesnt need to be absolute x if your squaring it.

the range of g is {y|y ≥ 0 ≤ y ≤ 1}
Correct
g of f isn't defined because domain is not real?!!

is that right
$$g=\sqrt{1-x^2}$$

$$f=x^2+2$$

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

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

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

Last edited:
It doesnt need to be absolute x if your squaring it.
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 $\{ y | 0 \leq y \leq 1\}$.

What are you missing

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 $x$.

so is this right?

the domain of g is {x||x| ≤ + 1}

indeed it is.

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

the domain of g is {x| x ≤ + 1}

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

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

no its not!

data no it isn't....

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

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 $1 < x^2 \Longrightarrow 1 - x^2 < 0$.

how come your latex looks so sexy

I was thinking x can go from -1,1 which is the same as |x| from 0 to 1
precisely. And $0 \leq |x| \leq 1$ is the same as $|x| \leq 1$ by the definition of an inner product.

how come your latex looks so sexy
Long physics lab reports. Believe me, it's not a good tradeoff

I think Im dernaged because I yearn those.

hey data....

can't i just say the domain of g is ....

{x|x ≤ + 1}

Once you spend two years of friday nights writing them for twelve hours in a row (like me!), you might not be so enamored with the idea anymore :rofl: