Composition of functions

[dotty4]

Hi,
For some reason I just don't understand what I am doing wrong with some of the questions and I have no idea how they came to the answers they did.

For example (this is my work)
#1
f(x)=10-x and g(x) = 20-x
(f o g) (x)
=10-(20-x)
=10-20+x
=-10 + x

(g o f) (x)
=20-(10-x)
=20-10+x
=x+10

Thank you

 

[masters]

Hi,
For some reason I just don't understand what I am doing wrong with some of the questions and I have no idea how they came to the answers they did.

For example (this is my work)
#1
f(x)=10-x and g(x) = 20-x
(f o g) (x)
=10-(20-x)
=10-20+x
=-10 + x

(g o f) (x)
=20-(10-x)
=20-10+x
=x+10

Thank you

Hi dotty4,

Who are "they" and what answers did "they" come up with?

Your compositions look fine.

 

[dotty4]

OOOPs!!!
I posted the ONLY one I got right hahahahha


Okay, here goes the other ones:

f(x)=3x and g(x)=x^2
(f o g)(x)
=(3x)^2
=(3x)(3x)

(g o f)(x)
=(3x)(3x)
=9+3x+3x+x^2
=x^2+6x+9


Example #2
f(x) = 9 and g(x) = 12-x
(f o g)(x)
=[f(12-x)]
=[9(12-x)]
Now this is what I don't know how they ended up with the answer
x=9

(g o f)(x)
[g(9)]
=12-x+9
=-x+12+9
Their answer is 13

(They refers to the text book)

I sort of understand but with that single number 9, do I add or multiply it?

 

[masters]

OOOPs!!!
I posted the ONLY one I got right hahahahha


Okay, here goes the other ones:

f(x)=3x and g(x)=x^2
(f o g)(x)
=(3x)^2
=(3x)(3x)

(g o f)(x)
=(3x)(3x)
=9+3x+3x+x^2
=x^2+6x+9

\(\displaystyle f(x)=3x\) and \(\displaystyle g(x)=x^2\)

\(\displaystyle f[g(x)]=f(x^2)=3(x^2)=3x^2\)

\(\displaystyle g[f(x)]=g(3x)=(3x)^2=9x^2\)

Example #2
f(x) = 9 and g(x) = 12-x
(f o g)(x)
=[f(12-x)]
=[9(12-x)]
Now this is what I don't know how they ended up with the answer
x=9

(g o f)(x)
[g(9)]
=12-x+9
=-x+12+9
Their answer is 13

\(\displaystyle f(x)=9\) and \(\displaystyle g(x)=12-x\)

There is no composition with f[g(x)] since f(x) is a constant.

\(\displaystyle g[f(x)]=g(9)=12-9=3\)

I don't see how "they" got 13 here.

 

[dotty4]

Thanks!!!
I might of copied it wrong.

I will be back on this thread, I need to pass this chapter *cry* I failed the last chapter.

 

[dotty4]

When you say it is a "constant" does that mean if its just a plain number it can't be made into anything?

 

[dotty4]

\(\displaystyle f(x)=9\) and \(\displaystyle g(x)=12-x\)

There is no composition with f[g(x)] since f(x) is a constant.

\(\displaystyle g[f(x)]=g(9)=12-9=3\)

I don't see how "they" got 13 here.

The text book says the answer is 9, what happened to the 12-x?

 

[masters]

\(\displaystyle f(x)=9\) and \(\displaystyle g(x)=12-x\)

There is no composition with f[g(x)] since f(x) is a constant.

Look at it this way. f(x) = 9 could be written as f(x) = 0x + 9.

To find f[g(x)] when g(x) = 12 - x, you would have f(12 - x) = 0(12 - x) + 9 = 9.

That's how they get their 9.

[quote:1kuve0tf]
\(\displaystyle g[f(x)]=g(9)=12-9=3\)

I don't see how "they" got 13 here.

The text book says the answer is 9, what happened to the 12-x?

[/quote:1kuve0tf]

My solution is correct. \(\displaystyle g[f(x)]=g(9)=12-9=3\)