Can anyone tell me where am I making mistakes here in composition of function problem

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The question is,

$$ f(x) = left| x-1 right| -1 , , x in [ -1,3] $$

$$ g(x) = 2 - left| x+1 right| , , x in [-2, 2] $$

Find $ f(g(x)) $

I did try, but the answer behind the text book has different domain of $f(g(x))$

Here's my work, can someone point out the mistake, if any?

https://www.scribd.com/document/384934557/My-work

Thank you :)

functions

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asked Jul 29 at 11:52

William

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up vote
1
down vote

favorite

The question is,

$$ f(x) = left| x-1 right| -1 , , x in [ -1,3] $$

$$ g(x) = 2 - left| x+1 right| , , x in [-2, 2] $$

Find $ f(g(x)) $

I did try, but the answer behind the text book has different domain of $f(g(x))$

Here's my work, can someone point out the mistake, if any?

https://www.scribd.com/document/384934557/My-work

Thank you :)

functions

share|cite|improve this question

asked Jul 29 at 11:52

William

731214

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

The question is,

$$ f(x) = left| x-1 right| -1 , , x in [ -1,3] $$

$$ g(x) = 2 - left| x+1 right| , , x in [-2, 2] $$

Find $ f(g(x)) $

I did try, but the answer behind the text book has different domain of $f(g(x))$

Here's my work, can someone point out the mistake, if any?

https://www.scribd.com/document/384934557/My-work

Thank you :)

functions

share|cite|improve this question

asked Jul 29 at 11:52

William

731214

The question is,

$$ f(x) = left| x-1 right| -1 , , x in [ -1,3] $$

$$ g(x) = 2 - left| x+1 right| , , x in [-2, 2] $$

Find $ f(g(x)) $

I did try, but the answer behind the text book has different domain of $f(g(x))$

Here's my work, can someone point out the mistake, if any?

https://www.scribd.com/document/384934557/My-work

Thank you :)

functions

share|cite|improve this question

asked Jul 29 at 11:52

William

731214

share|cite|improve this question

share|cite|improve this question

asked Jul 29 at 11:52

William

731214

asked Jul 29 at 11:52

William

731214

asked Jul 29 at 11:52

William

731214

731214

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add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote

$$g(x) = 2 - lvert x+1 rvert, qquad x in [-2, 2]$$
So
$$g(x) =
begincases
2+(x+1)=x+3, &xin[-2,-1]\
2-(x+1)=1-x, &xin(-1,2]
endcases$$
So

beginalign
f(x)
&= begincases
left| x+3-1 right| -1 =|x+2|-1, &xin[-2,-1]\
left| 1-x-1 right| -1 =|-x|-1, &xin(-1,2]
endcases \
&= begincases
|x+2|-1, &xin[-2,-1]\
|x|-1, &xin(-1,2]
endcases \
&= begincases
(x+2)-1, &xin[-2,-1]\
-x-1, &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&= begincases
x+1, &xin[-2,-1]\
-(x+1), &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&textOR\
f(x)&= begincases
x+1, &xin[-2,-1)\
-(x+1), &xin[-1,0)\
x-1,&xin[0,2]
endcases \
endalign

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answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait so my answer is correct?
  – William
  Jul 29 at 12:52
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

If $x in [-2,2]$ then $0 leq lvert x+1 rvert leq 3$
and therefore $-1 leq g(x) leq 2.$
Since these values of $g(x)$ are all within the domain of $f,$
I conclude that $f(g(x))$ is defined for every $x in [-2,2].$
The composition $f(g(x))$ cannot be defined for any $x$ that is not in the domain of $g,$ so no other points can be added to the domain.
The domain of $f(g(x))$ therefore is $[-2,2],$
exactly as you concluded.

Moreover, using an online tool to graph $f(g(x))$ confirms the rest of your definition of that function.

It might be helpful to see the book's answer to compare with yours.

share|cite|improve this answer

answered Jul 29 at 13:28

David K

48.1k340107

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

$$g(x) = 2 - lvert x+1 rvert, qquad x in [-2, 2]$$
So
$$g(x) =
begincases
2+(x+1)=x+3, &xin[-2,-1]\
2-(x+1)=1-x, &xin(-1,2]
endcases$$
So

beginalign
f(x)
&= begincases
left| x+3-1 right| -1 =|x+2|-1, &xin[-2,-1]\
left| 1-x-1 right| -1 =|-x|-1, &xin(-1,2]
endcases \
&= begincases
|x+2|-1, &xin[-2,-1]\
|x|-1, &xin(-1,2]
endcases \
&= begincases
(x+2)-1, &xin[-2,-1]\
-x-1, &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&= begincases
x+1, &xin[-2,-1]\
-(x+1), &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&textOR\
f(x)&= begincases
x+1, &xin[-2,-1)\
-(x+1), &xin[-1,0)\
x-1,&xin[0,2]
endcases \
endalign

share|cite|improve this answer

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait so my answer is correct?
  – William
  Jul 29 at 12:52
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

$$g(x) = 2 - lvert x+1 rvert, qquad x in [-2, 2]$$
So
$$g(x) =
begincases
2+(x+1)=x+3, &xin[-2,-1]\
2-(x+1)=1-x, &xin(-1,2]
endcases$$
So

beginalign
f(x)
&= begincases
left| x+3-1 right| -1 =|x+2|-1, &xin[-2,-1]\
left| 1-x-1 right| -1 =|-x|-1, &xin(-1,2]
endcases \
&= begincases
|x+2|-1, &xin[-2,-1]\
|x|-1, &xin(-1,2]
endcases \
&= begincases
(x+2)-1, &xin[-2,-1]\
-x-1, &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&= begincases
x+1, &xin[-2,-1]\
-(x+1), &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&textOR\
f(x)&= begincases
x+1, &xin[-2,-1)\
-(x+1), &xin[-1,0)\
x-1,&xin[0,2]
endcases \
endalign

share|cite|improve this answer

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait so my answer is correct?
  – William
  Jul 29 at 12:52
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

$$g(x) = 2 - lvert x+1 rvert, qquad x in [-2, 2]$$
So
$$g(x) =
begincases
2+(x+1)=x+3, &xin[-2,-1]\
2-(x+1)=1-x, &xin(-1,2]
endcases$$
So

beginalign
f(x)
&= begincases
left| x+3-1 right| -1 =|x+2|-1, &xin[-2,-1]\
left| 1-x-1 right| -1 =|-x|-1, &xin(-1,2]
endcases \
&= begincases
|x+2|-1, &xin[-2,-1]\
|x|-1, &xin(-1,2]
endcases \
&= begincases
(x+2)-1, &xin[-2,-1]\
-x-1, &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&= begincases
x+1, &xin[-2,-1]\
-(x+1), &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&textOR\
f(x)&= begincases
x+1, &xin[-2,-1)\
-(x+1), &xin[-1,0)\
x-1,&xin[0,2]
endcases \
endalign

share|cite|improve this answer

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

$$g(x) = 2 - lvert x+1 rvert, qquad x in [-2, 2]$$
So
$$g(x) =
begincases
2+(x+1)=x+3, &xin[-2,-1]\
2-(x+1)=1-x, &xin(-1,2]
endcases$$
So

beginalign
f(x)
&= begincases
left| x+3-1 right| -1 =|x+2|-1, &xin[-2,-1]\
left| 1-x-1 right| -1 =|-x|-1, &xin(-1,2]
endcases \
&= begincases
|x+2|-1, &xin[-2,-1]\
|x|-1, &xin(-1,2]
endcases \
&= begincases
(x+2)-1, &xin[-2,-1]\
-x-1, &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&= begincases
x+1, &xin[-2,-1]\
-(x+1), &xin(-1,0]\
x-1,&xin(0,2]
endcases \
&textOR\
f(x)&= begincases
x+1, &xin[-2,-1)\
-(x+1), &xin[-1,0)\
x-1,&xin[0,2]
endcases \
endalign

share|cite|improve this answer

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

answered Jul 29 at 12:30

Karn Watcharasupat

3,7992426

3,7992426

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait so my answer is correct?
  – William
  Jul 29 at 12:52
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Wait so my answer is correct?
  – William
  Jul 29 at 12:52
  
  

  
  
  
  

Wait so my answer is correct?
– William
Jul 29 at 12:52

Wait so my answer is correct?
– William
Jul 29 at 12:52

add a comment | 

up vote
0
down vote

If $x in [-2,2]$ then $0 leq lvert x+1 rvert leq 3$
and therefore $-1 leq g(x) leq 2.$
Since these values of $g(x)$ are all within the domain of $f,$
I conclude that $f(g(x))$ is defined for every $x in [-2,2].$
The composition $f(g(x))$ cannot be defined for any $x$ that is not in the domain of $g,$ so no other points can be added to the domain.
The domain of $f(g(x))$ therefore is $[-2,2],$
exactly as you concluded.

Moreover, using an online tool to graph $f(g(x))$ confirms the rest of your definition of that function.

It might be helpful to see the book's answer to compare with yours.

share|cite|improve this answer

answered Jul 29 at 13:28

David K

48.1k340107

add a comment | 

up vote
0
down vote

If $x in [-2,2]$ then $0 leq lvert x+1 rvert leq 3$
and therefore $-1 leq g(x) leq 2.$
Since these values of $g(x)$ are all within the domain of $f,$
I conclude that $f(g(x))$ is defined for every $x in [-2,2].$
The composition $f(g(x))$ cannot be defined for any $x$ that is not in the domain of $g,$ so no other points can be added to the domain.
The domain of $f(g(x))$ therefore is $[-2,2],$
exactly as you concluded.

Moreover, using an online tool to graph $f(g(x))$ confirms the rest of your definition of that function.

It might be helpful to see the book's answer to compare with yours.

share|cite|improve this answer

answered Jul 29 at 13:28

David K

48.1k340107

add a comment | 

up vote
0
down vote

up vote
0
down vote

If $x in [-2,2]$ then $0 leq lvert x+1 rvert leq 3$
and therefore $-1 leq g(x) leq 2.$
Since these values of $g(x)$ are all within the domain of $f,$
I conclude that $f(g(x))$ is defined for every $x in [-2,2].$
The composition $f(g(x))$ cannot be defined for any $x$ that is not in the domain of $g,$ so no other points can be added to the domain.
The domain of $f(g(x))$ therefore is $[-2,2],$
exactly as you concluded.

Moreover, using an online tool to graph $f(g(x))$ confirms the rest of your definition of that function.

It might be helpful to see the book's answer to compare with yours.

share|cite|improve this answer

answered Jul 29 at 13:28

David K

48.1k340107

If $x in [-2,2]$ then $0 leq lvert x+1 rvert leq 3$
and therefore $-1 leq g(x) leq 2.$
Since these values of $g(x)$ are all within the domain of $f,$
I conclude that $f(g(x))$ is defined for every $x in [-2,2].$
The composition $f(g(x))$ cannot be defined for any $x$ that is not in the domain of $g,$ so no other points can be added to the domain.
The domain of $f(g(x))$ therefore is $[-2,2],$
exactly as you concluded.

Moreover, using an online tool to graph $f(g(x))$ confirms the rest of your definition of that function.

It might be helpful to see the book's answer to compare with yours.

share|cite|improve this answer

answered Jul 29 at 13:28

David K

48.1k340107

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 13:28

David K

48.1k340107

answered Jul 29 at 13:28

David K

48.1k340107

answered Jul 29 at 13:28

David K

48.1k340107

48.1k340107

add a comment | 

add a comment | 

 

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