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Non-intuitive inverse function steps

2

$begingroup$

In my precalculus book (open source Stitz and Zeager) we are trying to find the inverse function for $x^2$ - 2x + 4. (page 390). The book has steps in here that just don't make sense to me. Here they are:

a.) $y = j(x)$

b.) $y = x^2 -2x + 4$, $x leq 1$

c.) $x = y^2 -2y + 4$, $x leq 1$

d.) $0 = y^2 -2y + 4 - x $

e.) $y = frac{2pmsqrt{(-2)^2 - 4(1)(4 - x)}}{2(1)}$

f.) $y = frac{2pmsqrt{4x - 12}}{2}$

and he continues on until he gets to step j

j.) 1$pmsqrt{x-3}$

My question is what is he doing in step d.)? To me it looks like he is taking the output (x in this case) and making it part of the function. From an algebra standpoint I guess this makes sense but from a function standpoint it has be bewildered. I wouldn't think you could do this (obviously I'm wrong but I need help understanding this intuitively).

Then in step e.) he is creating a brand new output y which to me would be a different output of x in step c.). This also looks very strange as I thought the point of creating the inverse function was to get the same x ouput as the x input of the original function. The book offers no explanation on these steps so I'm hoping someone can point me to some further reading on this technique. Does it have a name? Its just non-intuitive to me.

algebra-precalculus proof-explanation

share|cite|improve this question

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

asked Mar 22 at 22:14

maybedavemaybedave

866

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In step (c) the $xle1$ should be $yle1$.
  $endgroup$
  – John Wayland Bales
  Mar 22 at 22:31
  

  
  
  
  

add a comment | 

2

$begingroup$

In my precalculus book (open source Stitz and Zeager) we are trying to find the inverse function for $x^2$ - 2x + 4. (page 390). The book has steps in here that just don't make sense to me. Here they are:

a.) $y = j(x)$

b.) $y = x^2 -2x + 4$, $x leq 1$

c.) $x = y^2 -2y + 4$, $x leq 1$

d.) $0 = y^2 -2y + 4 - x $

e.) $y = frac{2pmsqrt{(-2)^2 - 4(1)(4 - x)}}{2(1)}$

f.) $y = frac{2pmsqrt{4x - 12}}{2}$

and he continues on until he gets to step j

j.) 1$pmsqrt{x-3}$

My question is what is he doing in step d.)? To me it looks like he is taking the output (x in this case) and making it part of the function. From an algebra standpoint I guess this makes sense but from a function standpoint it has be bewildered. I wouldn't think you could do this (obviously I'm wrong but I need help understanding this intuitively).

Then in step e.) he is creating a brand new output y which to me would be a different output of x in step c.). This also looks very strange as I thought the point of creating the inverse function was to get the same x ouput as the x input of the original function. The book offers no explanation on these steps so I'm hoping someone can point me to some further reading on this technique. Does it have a name? Its just non-intuitive to me.

algebra-precalculus proof-explanation

share|cite|improve this question

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

asked Mar 22 at 22:14

maybedavemaybedave

866

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In step (c) the $xle1$ should be $yle1$.
  $endgroup$
  – John Wayland Bales
  Mar 22 at 22:31
  

  
  
  
  

add a comment | 

$begingroup$

In my precalculus book (open source Stitz and Zeager) we are trying to find the inverse function for $x^2$ - 2x + 4. (page 390). The book has steps in here that just don't make sense to me. Here they are:

a.) $y = j(x)$

b.) $y = x^2 -2x + 4$, $x leq 1$

c.) $x = y^2 -2y + 4$, $x leq 1$

d.) $0 = y^2 -2y + 4 - x $

e.) $y = frac{2pmsqrt{(-2)^2 - 4(1)(4 - x)}}{2(1)}$

f.) $y = frac{2pmsqrt{4x - 12}}{2}$

and he continues on until he gets to step j

j.) 1$pmsqrt{x-3}$

My question is what is he doing in step d.)? To me it looks like he is taking the output (x in this case) and making it part of the function. From an algebra standpoint I guess this makes sense but from a function standpoint it has be bewildered. I wouldn't think you could do this (obviously I'm wrong but I need help understanding this intuitively).

Then in step e.) he is creating a brand new output y which to me would be a different output of x in step c.). This also looks very strange as I thought the point of creating the inverse function was to get the same x ouput as the x input of the original function. The book offers no explanation on these steps so I'm hoping someone can point me to some further reading on this technique. Does it have a name? Its just non-intuitive to me.

algebra-precalculus proof-explanation

share|cite|improve this question

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

asked Mar 22 at 22:14

maybedavemaybedave

866

$endgroup$

share|cite|improve this question

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

asked Mar 22 at 22:14

maybedavemaybedave

866

share|cite|improve this question

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

asked Mar 22 at 22:14

maybedavemaybedave

866

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

edited Mar 22 at 22:25

Maria Mazur

49.9k1361125

asked Mar 22 at 22:14

maybedavemaybedave

866

asked Mar 22 at 22:14

maybedavemaybedave

866

1 Answer
1

active

oldest

votes

2

$begingroup$

If we have $y=3x+4$ and we are trying to find an inverse we have to swap the role of dependent and independent variable so we get $x=3y+4$ and now we have to find out $y$ in this new formula, so $y={x-4over 3}$

The same is here: $x=y^2-2y+4$ and now we have to find out $y$. Since this is now quadrtics in $y$ we move everything on one side so: $$0=y^2-2y+(4-x)$$
So we get quadratic equation in $y$ with "parameter" $x$, and a solution is in $e$:

$$ y={2pmsqrt{4x-12}over 2} = 1pm sqrt{x-3}$$ Now since our $y<1$ (remember we swap the role of $x$ and $y$) we have $$y=1-sqrt{x-3}$$

share|cite|improve this answer

edited Mar 22 at 22:32

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, so it dawned on me when reading your reply that "y =" is not a new y in step e because when we use the quadratic formula we are solving for the variable in the first two terms (in this case y). So this now makes sense to me why we say "y =". But I'm still having issues understanding intuitively why we set c equal to 4 - x in step d. I get that it works but intuitively its not sitting well with me.
  $endgroup$
  – maybedave
  Mar 22 at 22:43
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we want to solve $$3 = y^2+3y +7 $$ then we have to move $3$ on right in order to get "standard" quadratic equation. The same is with $x$.
  $endgroup$
  – Maria Mazur
  Mar 22 at 22:46
  

  
  
  
  

add a comment | 

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2

$begingroup$

If we have $y=3x+4$ and we are trying to find an inverse we have to swap the role of dependent and independent variable so we get $x=3y+4$ and now we have to find out $y$ in this new formula, so $y={x-4over 3}$

The same is here: $x=y^2-2y+4$ and now we have to find out $y$. Since this is now quadrtics in $y$ we move everything on one side so: $$0=y^2-2y+(4-x)$$
So we get quadratic equation in $y$ with "parameter" $x$, and a solution is in $e$:

$$ y={2pmsqrt{4x-12}over 2} = 1pm sqrt{x-3}$$ Now since our $y<1$ (remember we swap the role of $x$ and $y$) we have $$y=1-sqrt{x-3}$$

share|cite|improve this answer

edited Mar 22 at 22:32

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, so it dawned on me when reading your reply that "y =" is not a new y in step e because when we use the quadratic formula we are solving for the variable in the first two terms (in this case y). So this now makes sense to me why we say "y =". But I'm still having issues understanding intuitively why we set c equal to 4 - x in step d. I get that it works but intuitively its not sitting well with me.
  $endgroup$
  – maybedave
  Mar 22 at 22:43
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we want to solve $$3 = y^2+3y +7 $$ then we have to move $3$ on right in order to get "standard" quadratic equation. The same is with $x$.
  $endgroup$
  – Maria Mazur
  Mar 22 at 22:46
  

  
  
  
  

add a comment | 

2

$begingroup$

If we have $y=3x+4$ and we are trying to find an inverse we have to swap the role of dependent and independent variable so we get $x=3y+4$ and now we have to find out $y$ in this new formula, so $y={x-4over 3}$

The same is here: $x=y^2-2y+4$ and now we have to find out $y$. Since this is now quadrtics in $y$ we move everything on one side so: $$0=y^2-2y+(4-x)$$
So we get quadratic equation in $y$ with "parameter" $x$, and a solution is in $e$:

$$ y={2pmsqrt{4x-12}over 2} = 1pm sqrt{x-3}$$ Now since our $y<1$ (remember we swap the role of $x$ and $y$) we have $$y=1-sqrt{x-3}$$

share|cite|improve this answer

edited Mar 22 at 22:32

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ok, so it dawned on me when reading your reply that "y =" is not a new y in step e because when we use the quadratic formula we are solving for the variable in the first two terms (in this case y). So this now makes sense to me why we say "y =". But I'm still having issues understanding intuitively why we set c equal to 4 - x in step d. I get that it works but intuitively its not sitting well with me.
  $endgroup$
  – maybedave
  Mar 22 at 22:43
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we want to solve $$3 = y^2+3y +7 $$ then we have to move $3$ on right in order to get "standard" quadratic equation. The same is with $x$.
  $endgroup$
  – Maria Mazur
  Mar 22 at 22:46
  

  
  
  
  

add a comment | 

$begingroup$

If we have $y=3x+4$ and we are trying to find an inverse we have to swap the role of dependent and independent variable so we get $x=3y+4$ and now we have to find out $y$ in this new formula, so $y={x-4over 3}$

The same is here: $x=y^2-2y+4$ and now we have to find out $y$. Since this is now quadrtics in $y$ we move everything on one side so: $$0=y^2-2y+(4-x)$$
So we get quadratic equation in $y$ with "parameter" $x$, and a solution is in $e$:

$$ y={2pmsqrt{4x-12}over 2} = 1pm sqrt{x-3}$$ Now since our $y<1$ (remember we swap the role of $x$ and $y$) we have $$y=1-sqrt{x-3}$$

share|cite|improve this answer

edited Mar 22 at 22:32

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125

$endgroup$

share|cite|improve this answer

edited Mar 22 at 22:32

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125

answered Mar 22 at 22:22

Maria MazurMaria Mazur

49.9k1361125