Hello there . I'm currently having a problem with a composite functions question . I know how to solve it , but have no idea as to why I should express y in terms of x before solving it . To get an idea , I've posted the question with a solution :



Why can't I just plug in x+4 in the expression x^2+8x+10 ? Why should I express the function f(x) = y first , and then express y in terms of x ?

This post has been edited by maximR: Jan 15 2012, 07:07 PM

Plugging in x+4 into the expression x^2 + 8x + 10 is wrong. k(x) is mapped into hk(x), if you plug x+4 in you'll be plugging in another k(x) into hk(x), making it become khk(x), and this will further complicate your answer as what you need now is finding h(x).

You have k(x), and now you know h(x) maps k(x) to hk(x). So now you need to find out h(x) from the final solution, like you need to find out an ingredient in a mixture. So what you have to do now is: you inverse it.

Let's say you know x+4 = 7. To find X you'll write so: x=7-4. The same thing applies here, the problem now is you do not know x+4 equals to what, so you put an unknown Y. So x+4 = y. And to find X you'll write x = y-4. It's just like you have mixed all ingredients into a mixture, and to find an ingredient you filter it to their individual ingredients by reversing the steps: you mix all ingredients into a mixture, now you filter the mixture to their respective ingredients. For this question, h(x) maps k(x) into hk(x), and now you reverse it by breaking hk(x).

This involves the knowledge on Inverse Function which is the last part of this chapter. Have you learnt it?

Sorry for the bad explanation, I'll just give a brief summary. Do ask your teacher for a more detailed explanation tomorrow.

This post has been edited by justarandomboy: Jan 15 2012, 07:25 PM

I thought you're pro, but you've showed me another example of no one is perfect ! Hahahaha

I've problems about this too, I just do the way let, and let .. without knowing the logic . Until before SPM, I asked here in LYN !

I got the logic .. actually you don't have to let, just that it will be very hard to figure out the answer .


So for your question, if I don't let ,

h(x+4) = x^2 + 8x + 10
h(x+4) = (x+4)^2 - 6
h(x)=x^2 - 6



This post has been edited by CallMeBin: Jan 15 2012, 10:10 PM

Mind explaining how you got the second step from the first? I don't get it.

(x+4)^2 = x^2 + 8x + 16


=)

What about the -6, it looks like it appears from nowhere. And where is +8x+10?

This post has been edited by justarandomboy: Jan 15 2012, 11:31 PM

You actually go backwards from the x^2+8x+10

Argh, a simple question.
Fyi, I'm a fully qualified UEC graduate with moderate results in Addmath I & II.

I can answer your question.

You see, because you need to find h(x).
And there is a function within the function, h(k(x)).
Which is also equal to h(x+4), because the question states so.

Therefore, to get h(x),
You need to define x+4 as y,

x+4=y
x=y-4

Done, I'm not really good at explaining with words. I tried my best.

I know....but

h(x+4) = x^2 + 8x + 10
h(x+4) = (x+4)^2 - 6
h(x)=x^2 - 6

I don't understand the steps above. How do you get the second step from the first?
If I do the first I'll get:
h(x+4) = (x+4)^2 + 8(x+4) + 10....
so....how do you get h(x+4) = (x+4)^2 - 6

This is the part I don't understand.....

I prefer the 'let' method, or the reverse function method.

hk(x) = (x^2 + 8x) + 10 --- A

Since
h(x+4) = (x^2 + 8x) + 16 - 6 --- B

Compare equation A and B

h(x+4) = (x+4)^2 -6

Change x+4 to x

h(x) = x^2 - 6

This post has been edited by VengenZ: Jan 15 2012, 11:43 PM

Its the same thing.

take k(x) for example.

h(x) = x or we say the function h(x), graphed as y = x

hence h(k(x)) = k(x)

Given k(x) = x + 2
we know hk(x) = (x + 2)

formalising an idea,
for an unknown function z(x),
if k(x) is a known function.

and zk(x) can be expressed as y = a + b(k(x)) + c or a function of k(x), for a,c, are known constants/parameters.
then z(x) is simply y = a + bx + c

which is how TS, used the "completing the Square method, to obtain the equation as a factor of (k(x))
and work out h(x). An alternative method from inversing the identity.

The first one

h(x+4) = x^2 + 8x + 10

The 2nd one

h(x+4) = x^2 + 8x + 16 - 6

Since they're the same equation which is h(x+4), so they must be the same .
(x+4)^2 get x^2 + 8x + 16,
so I must minus 6 to get 10


Oh god, if you really can't explain, don't explain, you will make the newbies confused .

Ok let's see
h[k(x)] , k(x) is the object
So, for convenience, you let y = k(x)
Therefore, h(y)

So continue ,
h(y) = x^2 + 8x + 10

As you can see, Left Equation, the object is y
While, Right equation the object is x
No sense right ? So you must make the object equal

So since k(x) = y = x+4
So now you have to get x instead of y to fulfill the same-object-criteria
Hence, x = y - 4

h(x) = x^2 + 8x + 10
Since, x = y-4
So substitute !

Oh gosh, couldn't believe that I still can explain !

Oh I get it now. thanks.

No problem

I didnt say the method was wrong

what method wrong?

Nvm.. I thought you quoted me coz something

Anyway..

Let me proof TS's technique (In hope that TS will understand).

h(x) = ax^2 + bx + c
k(x) = x + d

h[k(x)] = a[x+d]^2 + b(x+d) + c
hk(x) = a(x^2 + xd + d^2) + bx + bd + c
hk(x) = ax^2 + axd + ad^2 + bx + bd + c
hk(x) = ax^2 + x(ad+bx+bd) + ad^2 + c

Let B = ad+bx+bd and C = ad^2 + c

hk(x) = ax^2 + Bx + C

Now, we go backwards.

h[k(x)] = a[x+d]^2 + b(x+d) + c
h(x+d) = a[x+d]^2 + b(x+d) + c
h(x+d-d) = a[x+d-d]^2 + b(x+d-d) + c

Let x + d = y

h(y-d) = a(y-d)^2 + b(y-d) +c

Since y - d = x

h(x) = ax^2 + bx + c

(Q.E.D)

So in Ts's way of solving we have:

hk(x) = x^2 + 8x + 10
k(x) = x + 4
Let y = x + 4
x = y - 4

h(y) = (y - 4)^2 + 8(y - 4) + 10
h(y) = y^2 - 6

Hi . Never knew I'll get so many great responses . Anyway , yes , I finally got to the point that I can't understand the concept from all the reference books available , and also the textbooks . Since I'm currently tuition-less , I'll have to resort to asking on the internet , and I've gotten great responses .

From all of the posts above , I figured I'll simplify the main gist of the concept , at least for me :

We're finding h(x) , but if we map k(x) to the function x^2+8x+10 , I'll get the value of hk(x) instead and not h(x) . And to find x from k(x) , we'll have to inverse it , by letting y as the subject and then from there , express x in terms of y .

Since y is set to be equal to k(x) , which is x+4 , to solve for x , it will be x=y-4 .

But then when it comes to here my logic and reasoning doesn't understand this part : SINCE y=x+4 , and x=y-4 , why do we say its h(y) instead and not h(x) ? Is y=y-4 ?

Sorry for this shitty drawing (Hope you get the point)



h(y) = y^2 - 6 = h(x+4) = (x+4)^2 -6

This post has been edited by VengenZ: Jan 16 2012, 09:28 PM

Here are some exercises for your practice.

Suppose f(x) = (x-1)^2 - 1. Determine the following.
1. f(0) = ?
2. f(1) = ?
3. f(y) = ?
4. f(x^2) = ?
5. f(1/x) = ?
6. f(f(x)) = ?
7. Given g(x) = 1-x, what is f(g(x))?
8. Given g(x) = 1-x, what is g(f(x))?
9. Given f(g(x)) = x^2 + 4x + 3, and suppose g(x) = ax+b is linear. What are the constants a and b?
10. Given g(f(x)) = -2x^2 + 4x + 1, and suppose g(x) = ax+b is linear. What are the constants a and b?

If this is true , that we must express x in terms of y to get the same-object-criteria , then I get it . But shouldn't the right equation be named the image instead ?

---

Added on January 17, 2012, 6:33 pmI'm really tired , and after seeing it for a few times I still couldn't get it . I'm really too tired , back from merentas desa . I'll look at it again tomorrow and PM you about it .\

p/s : Pretty close . I'm about to get it .

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Added on January 17, 2012, 6:34 pmI need guidance for number 9 and 10 .

This post has been edited by maximR: Jan 17 2012, 06:37 PM

I really hope that you have attempted it by hard. Here is my solution, hope you can get some ideas after seeing through it. Good luck!

I finally got it after thinking through it many times for several days , with the help of you guys of course . I am proud to say that I got it without the help of a teacher , but a handful of people on the internet .

If we map the value , or expression of k(x) into the function hk(x) , then we're essentially mapping it to make it a much more complex function , but we're looking for h(x) . So , to get h(x) , we inverse k(x) to find x [ if we inverse k(x) , we're finding the initial object - x ] . Hence , after finding it , we can determine h(x) . I tried checking by substitution and finding the inverse in many ways , and found out that it is indeed the case . Then , we're writing it as h(y) because it's just an arbitrarily chosen letter , a place holder since we cannot have a function that is defined as h(x) = y-6 or etc ... So after solving it , we revert it back to h(x) - x-6 .

It's exhilarating finally understanding a concept . None of the seniors in my school managed to explain it to me .

Now , can we find the inverse of a composition of function , say gf(x) ? If yes then I'll get f(x) which will further reinforce my understanding of the concept .

This post has been edited by maximR: Jan 19 2012, 07:20 PM