# Need some Domain and Range Help

1. Jun 11, 2007

### MrRottenTreats

Hey here is the question has a few parts to it.

Let f(x)= (x-1)^1/2 and g(x)= 2x^2/x^2+1
a) State the domain and range of f using interval notation.
b) State the domain and range of g using interval notation. (Hint: At
x=0, g is 0. For x(cannot=0) it helps to write g(x)=2/1+1/x^2 and
then stetch a plot of g. Since x^2>0 it follows that g(x_ is always
below 2)
c) Find and simplify the expression for y(x)=(g o f)(x) and state its
domain and range in interval notation.
d) Find and simplify the expression for z(x)=(f o g)(x) and state its
domain and range in interval notation.

for a) i sketched a graph and made an x/y chart and subbed numbers in.
2 1
5 2
10 3

then got the domain to be: X E R | x >= 2
and the Rang YE R | y >= 1

domain: (2, infinity] ?
Range: (1, infitty] ?

and for b i did that same thing with the x/y chart
and i got domain: X E R
Range: Y E R | y >= 0

the hard part here is that i dont know how to put either of them in "interval notation"

and for that c) and d) i have no clue what y(x)=(g o f)(x) is asking me to do, and also z(x)=(f o g)(x)

some help would be great ! thanks.

2. Jun 11, 2007

### NateTG

Interval notation is something like:
(a,b) or [a,b) or (a,b] or [a,b]

I'm pretty sure that your range and domain aren't correct but you do understand interval notation.

Although you want to use the 'open' bracket for infinity:
$$(a,\infty)$$
and not
$$(a,\infty]$$
since the latter indicates that $\infty$ is part of the set.

3. Jun 11, 2007

### Pseudo Statistic

This should probably be in the homework help section.
Anyway, f o g means apply g to x first, then apply f to g.
So for example, say you have$$f(x) = 3x, g(x) = x^2, then f o g = f(g(x)) = f(x^2) = 3(x^2) = 3x^2$$
Obviously, in general, f o g =/= g o f.
OK, now for part a and b..
The domain of a function is the set of all numbers such that that function is defined; why is it that you say the domain of $$f(x) = \sqrt{x-1}$$ is $$[2,\infty]$$?
Also, what about what you found for the range? The range is the set of all points that the function hits, so to speak, whenever you give it its domain.
With the square root function, it's always greater than or equal to 0. Hence, to find its domain, you want to find all x such that $$\sqrt{x-1}\ge0$$ which you should have no problem solving.
Part B's domain is OK, but your range is off-- see below.
To write something like R, you write it as the interval $$(-\infty,\infty)$$, or $$x\ge2$$ as $$[2,\infty)$$ and so on and so forth.

About Part B's range-- $$g(x)= \frac{2x^2}{x^2+1}$$ can be written as $$g(x) = \frac{2x^2 + 2 - 2}{x^2 + 1} = 2 - \frac{2}{x^2 + 1}$$ (what you wrote is incorrect-- the way I'm seeing it, anyway).
From this, what can you gather from the behaviour of x, i.e. locally around 0 and as $$x\to \infty$$?

4. Jun 11, 2007

### MrRottenTreats

ok so for part a) i siad that cause when i plotted my points it started at x=2 , so thats why i had my domain as $$[2,\infty]$$

and for the rang it was the same thing, i just looked at my graph and tried to do it...

and also the same thing for b).. um im looking at them again and i cant really see where i went wrong with the domain and range.. maybe i graphed it wrong. What should the domain and range read?

now for c)
g o f = 2(x-1)^1/2)^2 / ((x-1)^1/2)^2+1

d)

f of g = ((2x^2/x^2+1)-1)^1/2

now what the heck do i do with these, if i got them done correctly.

thanks.

5. Jun 11, 2007

### NateTG

Well, does f(1) make sense? What about f(0)?