Real domain and range function to find all functions with nonzero x.
Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.
My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
algebra-precalculus functional-equations
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
asked Oct 14 '16 at 22:16
clache547
7611
add a comment |
Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.
My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
algebra-precalculus functional-equations
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
asked Oct 14 '16 at 22:16
clache547
7611
i did but it didn't really help
– clache547
Oct 14 '16 at 22:25
add a comment |
Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.
My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
algebra-precalculus functional-equations
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
asked Oct 14 '16 at 22:16
clache547
7611
Find all functions $f:mathbb R to mathbb R$ that satisfy $$ f(x) + 3 fleft( frac {x-1}{x} right) = 7x $$
for all nonzero $x$.
My thoughts so far are to plug in $(x - 1)/x$ for $x,$ but it always comes out so complicated, so I don't really know where to go from there.
algebra-precalculus functional-equations
algebra-precalculus functional-equations
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
asked Oct 14 '16 at 22:16
clache547
7611
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
asked Oct 14 '16 at 22:16
clache547
7611
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
edited Oct 14 '16 at 22:36
Simply Beautiful Art
50.2k578181
50.2k578181
asked Oct 14 '16 at 22:16
clache547
7611
asked Oct 14 '16 at 22:16
clache547
7611
asked Oct 14 '16 at 22:16
clache547
7611
7611
i did but it didn't really help
– clache547
Oct 14 '16 at 22:25
add a comment |
i did but it didn't really help
– clache547
Oct 14 '16 at 22:25
i did but it didn't really help
– clache547
Oct 14 '16 at 22:25
i did but it didn't really help
– clache547
Oct 14 '16 at 22:25
add a comment |
2 Answers
2
active
oldest
votes
$$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
$xtofrac{1}{1-x}$:
$$
fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
$$
$xto1-frac{1}{x}$:
$$
fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
$$
Hence, we have:
$$
begin{cases}
begin{align}
f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
end{align}
end{cases}
$$
Solve for $f(x)$:
$$
f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
$$
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
add a comment |
Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$
Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$
So we have three equations to work with:
begin{align}
f(x)+ 3f(1-tfrac {1}{x})= 7x,\
f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
end{align}
Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$
Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$
which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$
Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
$$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$
for $x ne 0,1.$
As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
for $x=1.$
(Also note that $f(x) = k$ for $x=0$)
answered Nov 28 '18 at 5:37
BobL
11
add a comment |
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2 Answers
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2 Answers
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$$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
$xtofrac{1}{1-x}$:
$$
fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
$$
$xto1-frac{1}{x}$:
$$
fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
$$
Hence, we have:
$$
begin{cases}
begin{align}
f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
end{align}
end{cases}
$$
Solve for $f(x)$:
$$
f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
$$
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
add a comment |
$$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
$xtofrac{1}{1-x}$:
$$
fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
$$
$xto1-frac{1}{x}$:
$$
fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
$$
Hence, we have:
$$
begin{cases}
begin{align}
f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
end{align}
end{cases}
$$
Solve for $f(x)$:
$$
f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
$$
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
add a comment |
$$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
$xtofrac{1}{1-x}$:
$$
fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
$$
$xto1-frac{1}{x}$:
$$
fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
$$
Hence, we have:
$$
begin{cases}
begin{align}
f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
end{align}
end{cases}
$$
Solve for $f(x)$:
$$
f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
$$
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
$$f(x) + 3 fleft( 1-frac {1}{x} right) = 7x$$
$xtofrac{1}{1-x}$:
$$
fleft(frac{1}{1-x}right)+3f(x)=7left(frac{1}{1-x}right)
$$
$xto1-frac{1}{x}$:
$$
fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)=7left(1-frac{1}{x}right)
$$
Hence, we have:
$$
begin{cases}
begin{align}
f(x)+3 fleft( 1-frac {1}{x} right)+0 &= 7x\
3f(x)+0+fleft(frac{1}{1-x}right)&=7left(frac{1}{1-x}right)\
0+fleft(1-frac{1}{x}right)+3fleft(frac{1}{1-x}right)&=7left(1-frac{1}{x}right)
end{align}
end{cases}
$$
Solve for $f(x)$:
$$
f(x)=-frac{3}{4}left(1-frac{1}{x}right)+frac{x}{4}+frac{9}{4}left(frac{1}{1-x}right)
$$
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
edited Oct 14 '16 at 23:05
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
answered Oct 14 '16 at 22:59
Kostiantyn Lapchevskyi
1,221414
1,221414
add a comment |
add a comment |
Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$
Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$
So we have three equations to work with:
begin{align}
f(x)+ 3f(1-tfrac {1}{x})= 7x,\
f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
end{align}
Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$
Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$
which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$
Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
$$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$
for $x ne 0,1.$
As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
for $x=1.$
(Also note that $f(x) = k$ for $x=0$)
answered Nov 28 '18 at 5:37
BobL
11
add a comment |
Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$
Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$
So we have three equations to work with:
begin{align}
f(x)+ 3f(1-tfrac {1}{x})= 7x,\
f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
end{align}
Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$
Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$
which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$
Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
$$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$
for $x ne 0,1.$
As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
for $x=1.$
(Also note that $f(x) = k$ for $x=0$)
answered Nov 28 '18 at 5:37
BobL
11
add a comment |
Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$
Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$
So we have three equations to work with:
begin{align}
f(x)+ 3f(1-tfrac {1}{x})= 7x,\
f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
end{align}
Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$
Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$
which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$
Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
$$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$
for $x ne 0,1.$
As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
for $x=1.$
(Also note that $f(x) = k$ for $x=0$)
answered Nov 28 '18 at 5:37
BobL
11
Substitute $x=tfrac{1}{1-x}$ to get $f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x).$
Substitute $x=1-tfrac{1}{x}$ to get $f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).$
So we have three equations to work with:
begin{align}
f(x)+ 3f(1-tfrac {1}{x})= 7x,\
f(tfrac{1}{1-x}) = 7(tfrac{1}{1-x})-3f(x),\
f(1-tfrac{1}{x}) = 7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}).
end{align}
Substituting the third equation into the first yields $$f(x)+ 3(7(1-tfrac{1}{x})- 3f(tfrac{1}{1-x}))= 7x,$$
which simplifies to :$$f(x)-9f(tfrac{1}{1-x}) = 7x-21(1-tfrac{1}{x}).$$
Then substitute in the second equation to yield $$f(x)-9(7(tfrac{1}{1-x})-3f(x)) = 7x-21(1-tfrac{1}{x}).$$
which expands and simplifies to $$28f(x)=7x-21+tfrac{21}{x}+tfrac{63}{1-x}.$$
Divide both sides by 28, put the right side over a common denominator, and multiply both the numerator and denominator on the right by $-1$ to yield
$$f(x) = frac{x^3-4x^2-3x-3}{4x(x-1)}$$
for $x ne 0,1.$
As the question asked for solutions to all nonzero x, we can substitute back into the original equation and find that $$f(x)= 7-3k$$
for $x=1.$
(Also note that $f(x) = k$ for $x=0$)
answered Nov 28 '18 at 5:37
BobL
11
edited Nov 29 '18 at 20:22
answered Nov 28 '18 at 5:37
BobL
11
answered Nov 28 '18 at 5:37
BobL
11
answered Nov 28 '18 at 5:37
BobL
11
11
add a comment |
add a comment |
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i did but it didn't really help
– clache547
Oct 14 '16 at 22:25