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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Zahta on December 28, 2011, 04:04:36 pm

Title: Composite function question Unit3 &4
Post by: Zahta on December 28, 2011, 04:04:36 pm: hey guys i need help with a composite function question , would appreciate any help thankyou:

f(x)=2x+1 ,g(x)= | x |
i need to find g(f(x) .
i get lost at the stage 2(|x | +1)
what do i do next .

Thankyou guys means alot if get any help
Title: Re: Composite function question Unit3 &4
Post by: Phy124 on December 28, 2011, 04:07:49 pm: g(f(x)) means you are putting the function f(x) in replacement of x within the function g(x)

therefore, g(f(x)) = |2x + 1| i.e. 2x + 1 has replaced x

edit: You may want to watch Composition of Functions for more help. Alternatively, you can ask related questions here, everyone is more than happy to help.
Title: Re: Composite function question Unit3 &4
Post by: Zahta on December 28, 2011, 04:33:59 pm: i understand compostion functions , just the |x | bit , but thankyou
Title: Re: Composite function question Unit3 &4
Post by: b^3 on December 28, 2011, 04:40:29 pm: The |x| bit is just a modulus. It makes anything that is neagtive into a positive, e.g. |-5|=5 and |5|=5. For |f(x)| it has the effect of flipping everything that is below the x-axis in the x-axis so that it becomes positive. For f(|x|) it has the effect of flipping the positive x-axis part of the graph in the y-axis (while retaining the positive part of the x-axis).

e.g. let f(x)=x-1, g(x)=|x| then g(f(x))=
y=|x-1|, i.e. y=x-1, for x<1, the graph is flipped in the x-axis
https://www.desmos.com/calculator/64744e5607

e.g. let f(x)=x-1, g(x)=|x| then f(g(x))=
y=|x|-1, the graph of y=x-1 for x>=1 is flipped in the y-axis (retaining the original graph for x>1)
https://www.desmos.com/calculator/f007a3a2e8

EDIT: If you want to draw it manually using hybrid functions, then you can do this. I'll use the one in your question.
As we know that whenever there is a negative in the modulus it will be made into a positive, so we need to find when the graph is negative. So $2x+1< 0$
$x\leq \frac{-1}{2}$
So for $x< \frac{-1}{2}$ the graph of f(x) will be flipped in the x-axis, that is it becomes -f(x)

Now using the hybird function
$y=\begin{cases}<br />2x+1 & \text{ if } x\geq -\frac{1}{2}\\ <br />-2x-1 & \text{ if } x< -\frac{1}{2} <br />\end{cases}$

https://www.desmos.com/calculator/8e48f5c59d

Hope that helps.

EDIT2: fixed the 0 and 1
Title: Re: Composite function question Unit3 &4
Post by: Zahta on December 28, 2011, 05:35:12 pm: b^3 you are a legend , thank you so very much. Helped alot can i ask another question

4. For the functions h:R\{0}arrow R,h(x)= 1/x² and g:R⁺arrow R,g(x)=3x+2,Find:

a) hºg(state rule & domain ) b)gºh ( state rule and domain)

I've stated the rule but confused about the domain.
Title: Re: Composite function question Unit3 &4
Post by: b^3 on December 28, 2011, 06:09:16 pm: This is how I remember how composite functions are defined.
'The range of the second is equal to or a subset of the domain of the first'.

I.e. If we had fog(x) then Ran g $\subseteq$ dom f i.e. the range of g(x) fits inside the domain of f(x)
I.e. If we had gof(x) then Ran f $\subseteq$ dom g i.e. the range of f(x) fits inside the domain of g(x)

For hog(x), the range of g(x) must be equal to or a subset of the domain of h(x).
So domain of h(x)=R\{0}
We need to restrict the domain of g(x) so that the range of g(x) is within/equal to R\{0}.
Currently the range of g(x) is $[2,\infty)$ but we want it to be equal/a subest of R\{0}. Currently the range of g(x) is R⁺, so we don't need to restrict it. Now the domain of hog(x) will be the domain of g(x). i.e. the domain of hog(x)=R⁺

So for goh(x), the range of h(x) must be equal to or within the domain of g(x)
So domain of g(x)=R⁺
We need to restrict the domain of h(x) so that the range of h(x) is within/equal to R⁺.
Currently the range of h(x) is R⁺, so we don't need to restrict it since it is equal to domain g(x)
Now the domain of goh(x) will be the domain of h(x). i.e. domain of goh(x)=R⁺

So for those two, you didn't need to restrict the domain because the range of the second was already equal/a subset of the domain of the first. Other cases you will need to restrict it.

After all that (sorry) I think the one bit of information that you needed was this: Once the domain has been restricted for the composite function, the domain of the composite function will be the domain of the second function.

EDIT: To make it simpler to see, most people draw up a table like so

| Dom | Ran |
g(x) R⁺ $[2,\infty)$
h(x) R\{0} R⁺

Then work out the retriction from there.
Title: Re: Composite function question Unit3 &4
Post by: Zahta on January 09, 2012, 07:14:35 am: i thought the domian for hog was the domain of g(x) which is g:R+arrow R, why is it just R+ and not g:R+arrow R.
Title: Re: Composite function question Unit3 &4
Post by: b^3 on January 09, 2012, 11:18:36 am: Is this the part you mean?
Quote from: b^3 on December 28, 2011, 06:09:16 pm
Currently the range of g(x) is $[2,\infty)$ but we want it to be equal/a subest of R\{0}. Currently the range of g(x) is R⁺, so we don't need to restrict it. Now the domain of hog(x) will be the domain of g(x). i.e. the domain of hog(x)=R⁺
The g:R+ arrow R means that 'R+ is mapped onto R' (If I'm not mistaken, dc or someone might pick up on if thats entierly correct or not). But usually we write the g:R+ arrow R when we write the funciton out the proper way at the end. I'm pretty sure that when you are doing all the working throughout you just need the R+ (at VCE level anyway).

I'm not entierly sure if I answered you're question or not.