1. ## Composite functions

Hi, I'm trying to understand the example the book gives:

For the functions $f(x)=x^2-1, x \in R,$ and $g(x) = \sqrt{x},x \geq 0$, state why $g(f(x))$ is not defined.

Range of $f \not\subseteq$ domain of g.
Therefore, g(f(x)) is not defined.

How can it not be defined? I've drawn graphs of $y = \sqrt{x^2-1}$ and they look perfectly reasonable. What is going on? Thanks

2. Originally Posted by DivideBy0
Hi, I'm trying to understand the example the book gives:

For the functions $f(x)=x^2-1, x \in R,$ and $g(x) = \sqrt{x},x \geq 0$, state why $g(f(x))$ is not defined.

Range of $f \not\subseteq$ domain of g.
Therefore, g(f(x)) is not defined.

How can it not be defined? I've drawn graphs of $y = \sqrt{x^2-1}$ and they look perfectly reasonable. What is going on? Thanks

It is defined on its domain, and its domain is $f^{-1} (D_g)$, where $D_g$ is the domain of $g$

ZB

3. sorry i don't think i'm getting it

4. For the functions and , state why is not defined.
Actually, g[f(x)] is defined on the entire domain of g: zero to infinity EXCEPT for values between zero and one.

5. Originally Posted by DivideBy0
Hi, I'm trying to understand the example the book gives:

For the functions $f(x)=x^2-1, x \in R,$ and $g(x) = \sqrt{x},x \geq 0$, state why $g(f(x))$ is not defined.

Range of $f \not\subseteq$ domain of g.
Therefore, g(f(x)) is not defined.

How can it not be defined? I've drawn graphs of $y = \sqrt{x^2-1}$ and they look perfectly reasonable. What is going on? Thanks
Say $f(x)$ is defined on domain $A$. And $g(x)$ is defined on domain $B$. Suppose we can compose them $f(g(x))$ in order for this expression to make sense: first we need that $x\in B$ because we evaluate at $B$ first; then we want $g(x)\in A$ because we evaluate at $f(x)$ second. So the rule is for $f(g(x))$ to be defined we require that the range (output values) of $g(x)$ to be contained in $A$, the domain $f(x)$.

6. Thanks everyone,

So... the composition is 'undefined', even though there are only a few points where it is actually undefined out of infinity. Well I guess $|x| <1$ does contain infinity points... but that argument seems a bit strained.

So if $f(x) = x^2-1, \ x \in [0, \infty)$, then g(f(x)) would be defined, right?

7. So if , then g(f(x)) would be defined, right?
Not quite. You should be (and probably are) thinking of

$f(x) = x^2-1, f(x)\in [0,\infty)$

8. Yep, that was what I was 'trying' to think of :P

I have another question, is it always true that $ran f(g(x)) = ranf$?

9. I have another question, is it always true that ?
No.

$dom \ f \circ g=ran\ g$?

(My book says it's $dom \ f \circ g = dom \ g$, but I don't believe this)

Sine $ran\ f \subseteq dom \ g$, so i think you can only choose from the $ran\ f$ subset for the domain of the $dom \ f \circ g$.

Then $ran \ f \circ g$ can be determined by only graphing the values $x \in ran \ g$?

I think I'm finally getting my head around this, thanks so much for all your help.

11. ?
no

(My book says it's , but I don't believe this)

Sine , so i think you can only choose from the subset for the domain of the .
Your book is right. A visualisation that might help you is an g-machine, into which you put in a number that is inside dom g, and an f-machine, into which you put a number which is inside dom f. The machine will then print out a number that is inside its range. f o g is the machine made when you attach the input of f to the output of g. It will only work if ran g $\subseteq$ dom f (you wrote this the wrong way round in your post). Otherwise g may print out a number that f can't handle.
Since ran g $\subseteq$dom f, the new machine can handle any input in dom g.

what confused and irritated me when I was learning this is: Why don't we make a machine that works even if ran g $\not\subseteq$ dom f by making a smaller domain so that the machine only receives input that will produce a g(x) that is inside dom f?
The answer is that we do, but we don't call it defined.

I think I'm finally getting my head around this, thanks so much for all your help.