Checking if the function is onto
$begingroup$
Consider a function $f$ defined over $R to R$
$y=x^2+2x$
Well it's inverse comes out to be
$f^{-1}(x)=sqrt {x+1} -1$
Well, this seems to be undefined for all values of co-domain $lt -1$
Am I correct in my reasoning?
functions
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
$endgroup$
add a comment |
$begingroup$
Consider a function $f$ defined over $R to R$
$y=x^2+2x$
Well it's inverse comes out to be
$f^{-1}(x)=sqrt {x+1} -1$
Well, this seems to be undefined for all values of co-domain $lt -1$
Am I correct in my reasoning?
functions
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
$endgroup$
$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06
add a comment |
$begingroup$
Consider a function $f$ defined over $R to R$
$y=x^2+2x$
Well it's inverse comes out to be
$f^{-1}(x)=sqrt {x+1} -1$
Well, this seems to be undefined for all values of co-domain $lt -1$
Am I correct in my reasoning?
functions
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
$endgroup$
Consider a function $f$ defined over $R to R$
$y=x^2+2x$
Well it's inverse comes out to be
$f^{-1}(x)=sqrt {x+1} -1$
Well, this seems to be undefined for all values of co-domain $lt -1$
Am I correct in my reasoning?
functions
functions
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
asked Dec 25 '18 at 11:57
user3767495user3767495
4028
4028
$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06
add a comment |
$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06
$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06
$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06
add a comment |
2 Answers
2
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$begingroup$
We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.
Can you proceed?
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.
Can you proceed?
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.
Can you proceed?
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.
Can you proceed?
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
$endgroup$
We have $f (x)=(x+1)^2-1 ge -1$ for all $x $. Hence $f (mathbb R)=[-1, infty) $.
Can you proceed?
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
answered Dec 25 '18 at 12:03
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
$begingroup$
If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
$endgroup$
add a comment |
$begingroup$
If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
$endgroup$
add a comment |
$begingroup$
If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
$endgroup$
If the function is onto, one would expect a reverse for any value. Otherwise, it would be into.
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
answered Dec 25 '18 at 12:01
wendy.kriegerwendy.krieger
5,85511427
5,85511427
add a comment |
add a comment |
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$begingroup$
If your reasoning concludes "Therefore this function is NOT 'onto'" then, yes, you are correct. You could also do this problem by 'completing the square': $y= x^2+ 2x= x^2+ 2x+ 1- 1= (x+ 1)^2- 1$. When x= -1, y= -1. For any x other than -1, $(x+ 1)^2$ is larger than 0 so y is larger than -1. There is no x such that y is, say, -2 so this function is NOT 'onto'.
$endgroup$
– user247327
Dec 25 '18 at 12:06