Is this equation solvable (by hand) and what do you call this equation?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Is this equation solvable (by hand) and what do you call this equation?



$a = x - min(b, x, c)$



$a, b, c$ are known. We need to solve for $x$.



If it is, and I can get guidance as to what this is called, and where there are resources to find out how to solve, this, I'm happy to do this.






maxima-minima







share|cite|improve this question























  • Does * mean the product?
    – mrs
    Sep 4 at 12:33






  • 1




    You simply test each condition: $a=x-bx textif bx<c$ and $a=x-c$ if $bx>c$. For equality choose either of the condtions. Solve for $x$ and test which condition holds.
    – player100
    Sep 4 at 12:58











  • @ResidentDementor yes, * is product.
    – RyanH
    Sep 4 at 13:14










  • Of course $x-min(bx,c) = max(x-bx,x-c)$
    – GEdgar
    Sep 4 at 14:40














up vote
0
down vote

favorite












Is this equation solvable (by hand) and what do you call this equation?



$a = x - min(b, x, c)$



$a, b, c$ are known. We need to solve for $x$.



If it is, and I can get guidance as to what this is called, and where there are resources to find out how to solve, this, I'm happy to do this.






maxima-minima







share|cite|improve this question























  • Does * mean the product?
    – mrs
    Sep 4 at 12:33






  • 1




    You simply test each condition: $a=x-bx textif bx<c$ and $a=x-c$ if $bx>c$. For equality choose either of the condtions. Solve for $x$ and test which condition holds.
    – player100
    Sep 4 at 12:58











  • @ResidentDementor yes, * is product.
    – RyanH
    Sep 4 at 13:14










  • Of course $x-min(bx,c) = max(x-bx,x-c)$
    – GEdgar
    Sep 4 at 14:40












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is this equation solvable (by hand) and what do you call this equation?



$a = x - min(b, x, c)$



$a, b, c$ are known. We need to solve for $x$.



If it is, and I can get guidance as to what this is called, and where there are resources to find out how to solve, this, I'm happy to do this.






maxima-minima







share|cite|improve this question















Is this equation solvable (by hand) and what do you call this equation?



$a = x - min(b, x, c)$



$a, b, c$ are known. We need to solve for $x$.



If it is, and I can get guidance as to what this is called, and where there are resources to find out how to solve, this, I'm happy to do this.






maxima-minima




maxima-minima






share|cite|improve this question















share|cite|improve this question













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edited Sep 4 at 12:34









Bernard

112k635104




112k635104










asked Sep 4 at 12:31









RyanH

1033




1033











  • Does * mean the product?
    – mrs
    Sep 4 at 12:33






  • 1




    You simply test each condition: $a=x-bx textif bx<c$ and $a=x-c$ if $bx>c$. For equality choose either of the condtions. Solve for $x$ and test which condition holds.
    – player100
    Sep 4 at 12:58











  • @ResidentDementor yes, * is product.
    – RyanH
    Sep 4 at 13:14










  • Of course $x-min(bx,c) = max(x-bx,x-c)$
    – GEdgar
    Sep 4 at 14:40
















  • Does * mean the product?
    – mrs
    Sep 4 at 12:33






  • 1




    You simply test each condition: $a=x-bx textif bx<c$ and $a=x-c$ if $bx>c$. For equality choose either of the condtions. Solve for $x$ and test which condition holds.
    – player100
    Sep 4 at 12:58











  • @ResidentDementor yes, * is product.
    – RyanH
    Sep 4 at 13:14










  • Of course $x-min(bx,c) = max(x-bx,x-c)$
    – GEdgar
    Sep 4 at 14:40















Does * mean the product?
– mrs
Sep 4 at 12:33




Does * mean the product?
– mrs
Sep 4 at 12:33




1




1




You simply test each condition: $a=x-bx textif bx<c$ and $a=x-c$ if $bx>c$. For equality choose either of the condtions. Solve for $x$ and test which condition holds.
– player100
Sep 4 at 12:58





You simply test each condition: $a=x-bx textif bx<c$ and $a=x-c$ if $bx>c$. For equality choose either of the condtions. Solve for $x$ and test which condition holds.
– player100
Sep 4 at 12:58













@ResidentDementor yes, * is product.
– RyanH
Sep 4 at 13:14




@ResidentDementor yes, * is product.
– RyanH
Sep 4 at 13:14












Of course $x-min(bx,c) = max(x-bx,x-c)$
– GEdgar
Sep 4 at 14:40




Of course $x-min(bx,c) = max(x-bx,x-c)$
– GEdgar
Sep 4 at 14:40










1 Answer
1






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up vote
1
down vote



accepted










Expanding on the comment by player100:



The function $f(x) = x-min(bx,c) $ has two regions:



$f(x) = (1-b)x quad textif bx le c$



$f(x) = x-c quad textif bx gt c$



Setting $f(x) = a$ then we have



$x = fraca1-b quad textif fracab1-b le c$



$x = c+a quad textif b(c+a) gt c$



Note that we can re-arrange the condition $b(c+a) gt b$ as $ab gt (1-b)c$, so the two conditions are mutually exclusive if $1-b > 0$ i.e. $b < 1$. So if $b< 1$ then the equation $f(x)=a$ has exactly one solution.



However, if $b > 1$ then the equation $f(x)=a$ will have two solutions if $ab > (1-b)c$, one solution if $ab = (1-b)c$ and no solutions if $ab < (1-b)c$.






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  • Thank you very much for your help and solution!
    – RyanH
    Sep 6 at 10:15










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Expanding on the comment by player100:



The function $f(x) = x-min(bx,c) $ has two regions:



$f(x) = (1-b)x quad textif bx le c$



$f(x) = x-c quad textif bx gt c$



Setting $f(x) = a$ then we have



$x = fraca1-b quad textif fracab1-b le c$



$x = c+a quad textif b(c+a) gt c$



Note that we can re-arrange the condition $b(c+a) gt b$ as $ab gt (1-b)c$, so the two conditions are mutually exclusive if $1-b > 0$ i.e. $b < 1$. So if $b< 1$ then the equation $f(x)=a$ has exactly one solution.



However, if $b > 1$ then the equation $f(x)=a$ will have two solutions if $ab > (1-b)c$, one solution if $ab = (1-b)c$ and no solutions if $ab < (1-b)c$.






share|cite|improve this answer




















  • Thank you very much for your help and solution!
    – RyanH
    Sep 6 at 10:15














up vote
1
down vote



accepted










Expanding on the comment by player100:



The function $f(x) = x-min(bx,c) $ has two regions:



$f(x) = (1-b)x quad textif bx le c$



$f(x) = x-c quad textif bx gt c$



Setting $f(x) = a$ then we have



$x = fraca1-b quad textif fracab1-b le c$



$x = c+a quad textif b(c+a) gt c$



Note that we can re-arrange the condition $b(c+a) gt b$ as $ab gt (1-b)c$, so the two conditions are mutually exclusive if $1-b > 0$ i.e. $b < 1$. So if $b< 1$ then the equation $f(x)=a$ has exactly one solution.



However, if $b > 1$ then the equation $f(x)=a$ will have two solutions if $ab > (1-b)c$, one solution if $ab = (1-b)c$ and no solutions if $ab < (1-b)c$.






share|cite|improve this answer




















  • Thank you very much for your help and solution!
    – RyanH
    Sep 6 at 10:15












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Expanding on the comment by player100:



The function $f(x) = x-min(bx,c) $ has two regions:



$f(x) = (1-b)x quad textif bx le c$



$f(x) = x-c quad textif bx gt c$



Setting $f(x) = a$ then we have



$x = fraca1-b quad textif fracab1-b le c$



$x = c+a quad textif b(c+a) gt c$



Note that we can re-arrange the condition $b(c+a) gt b$ as $ab gt (1-b)c$, so the two conditions are mutually exclusive if $1-b > 0$ i.e. $b < 1$. So if $b< 1$ then the equation $f(x)=a$ has exactly one solution.



However, if $b > 1$ then the equation $f(x)=a$ will have two solutions if $ab > (1-b)c$, one solution if $ab = (1-b)c$ and no solutions if $ab < (1-b)c$.






share|cite|improve this answer












Expanding on the comment by player100:



The function $f(x) = x-min(bx,c) $ has two regions:



$f(x) = (1-b)x quad textif bx le c$



$f(x) = x-c quad textif bx gt c$



Setting $f(x) = a$ then we have



$x = fraca1-b quad textif fracab1-b le c$



$x = c+a quad textif b(c+a) gt c$



Note that we can re-arrange the condition $b(c+a) gt b$ as $ab gt (1-b)c$, so the two conditions are mutually exclusive if $1-b > 0$ i.e. $b < 1$. So if $b< 1$ then the equation $f(x)=a$ has exactly one solution.



However, if $b > 1$ then the equation $f(x)=a$ will have two solutions if $ab > (1-b)c$, one solution if $ab = (1-b)c$ and no solutions if $ab < (1-b)c$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 4 at 13:41









gandalf61

6,159522




6,159522











  • Thank you very much for your help and solution!
    – RyanH
    Sep 6 at 10:15
















  • Thank you very much for your help and solution!
    – RyanH
    Sep 6 at 10:15















Thank you very much for your help and solution!
– RyanH
Sep 6 at 10:15




Thank you very much for your help and solution!
– RyanH
Sep 6 at 10:15

















 

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