calculate $f(500)=?$ and give an example of a function that meets the given conditions The Next CEO of Stack OverflowA question about Darboux functionsGive an example of a function such thatGive an example of a function $h$ that is discontinuous at every point of $[0,1]$, but with $|h|$ continuous on $[0,1]$Examples of function the following three conditionsGive an example of a continuous function with this propertyHow to find a function that satisfies 2 conditionsExample of a function that converges to 0 pointwise but integral is 3/2?Darboux function and its absolute value being continuousGive an example of a function $f:mathbbR to mathbbR$ that satisfies all three conditionsProve that the polynomial function is increasing in $ (-epsilon ,epsilon )$
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calculate $f(500)=?$ and give an example of a function that meets the given conditions
The Next CEO of Stack OverflowA question about Darboux functionsGive an example of a function such thatGive an example of a function $h$ that is discontinuous at every point of $[0,1]$, but with $|h|$ continuous on $[0,1]$Examples of function the following three conditionsGive an example of a continuous function with this propertyHow to find a function that satisfies 2 conditionsExample of a function that converges to 0 pointwise but integral is 3/2?Darboux function and its absolute value being continuousGive an example of a function $f:mathbbR to mathbbR$ that satisfies all three conditionsProve that the polynomial function is increasing in $ (-epsilon ,epsilon )$
$begingroup$
$f : mathbbR to mathbbR$
$forall_xin mathbbR f(x) * f(f(x))=1 (*) \
f(1000)=999$
calculate $f(500)=?$
So $f(1000)*f(f(1000))=1 to
999*f(999)=1 to
f(999)= frac1999$
Darboux.
$exists x_0 in (999,1000)$
With (*)
$ x=x_0$
$f(x_0)*f(f(x_0))=1 to $
$500*f(500)=1 to $
$f(500)=frac1500$
I don't know how to give an example of a function that meets the given conditions.
analysis functions continuity
asked Mar 20 at 9:45
mona1lisamona1lisa
727
$endgroup$
add a comment |
$begingroup$
$f : mathbbR to mathbbR$
$forall_xin mathbbR f(x) * f(f(x))=1 (*) \
f(1000)=999$
calculate $f(500)=?$
So $f(1000)*f(f(1000))=1 to
999*f(999)=1 to
f(999)= frac1999$
Darboux.
$exists x_0 in (999,1000)$
With (*)
$ x=x_0$
$f(x_0)*f(f(x_0))=1 to $
$500*f(500)=1 to $
$f(500)=frac1500$
I don't know how to give an example of a function that meets the given conditions.
analysis functions continuity
asked Mar 20 at 9:45
mona1lisamona1lisa
727
$endgroup$
add a comment |
$begingroup$
$f : mathbbR to mathbbR$
$forall_xin mathbbR f(x) * f(f(x))=1 (*) \
f(1000)=999$
calculate $f(500)=?$
So $f(1000)*f(f(1000))=1 to
999*f(999)=1 to
f(999)= frac1999$
Darboux.
$exists x_0 in (999,1000)$
With (*)
$ x=x_0$
$f(x_0)*f(f(x_0))=1 to $
$500*f(500)=1 to $
$f(500)=frac1500$
I don't know how to give an example of a function that meets the given conditions.
analysis functions continuity
asked Mar 20 at 9:45
mona1lisamona1lisa
727
$endgroup$
$f : mathbbR to mathbbR$
$forall_xin mathbbR f(x) * f(f(x))=1 (*) \
f(1000)=999$
calculate $f(500)=?$
So $f(1000)*f(f(1000))=1 to
999*f(999)=1 to
f(999)= frac1999$
Darboux.
$exists x_0 in (999,1000)$
With (*)
$ x=x_0$
$f(x_0)*f(f(x_0))=1 to $
$500*f(500)=1 to $
$f(500)=frac1500$
I don't know how to give an example of a function that meets the given conditions.
analysis functions continuity
analysis functions continuity
asked Mar 20 at 9:45
mona1lisamona1lisa
727
asked Mar 20 at 9:45
mona1lisamona1lisa
727
asked Mar 20 at 9:45
mona1lisamona1lisa
727
asked Mar 20 at 9:45
mona1lisamona1lisa
727
asked Mar 20 at 9:45
mona1lisamona1lisa
727
727
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any $y$ in the image of $f$ will have $y=f(x)$ for some $x in mathbb R$ and then satisfy $y times f(y)=1$ so $f(y)=frac1y$ and thus $fleft(frac1yright)=y$. Let's call $f$'s image $Y=y in mathbb R mid exists x in mathbb R: y=f(x)$
You now know that $0$ cannot be in $Y$. Since $f(1000)=999$, you also know $1000$ and $frac11000$ cannot be in $Y$ and that $999$ and $frac1999$ must be in $Y$. Any such $f$ and $Y$ which meets these conditions with $f(y)=frac1y$ for $y in Y$ will be a solution. So $f(500)$ can potentially take any value apart from $0$ or $1000$ or $frac11000$ or $500$
To construct an example, suppose you want $f(500)=k not in left0 , 1000,frac11000,500 right$. Then consider
$$
f(x) =
begincases
999 &quadtextif x= 1000 text or x= tfrac1999\
tfrac1999 &quadtextif x= 999\
tfrac1k &quadtextif x= k\
k &quadtextotherwise. \
endcases$$
There will be many other possible examples
answered Mar 20 at 17:27
HenryHenry
101k482169
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Any $y$ in the image of $f$ will have $y=f(x)$ for some $x in mathbb R$ and then satisfy $y times f(y)=1$ so $f(y)=frac1y$ and thus $fleft(frac1yright)=y$. Let's call $f$'s image $Y=y in mathbb R mid exists x in mathbb R: y=f(x)$
You now know that $0$ cannot be in $Y$. Since $f(1000)=999$, you also know $1000$ and $frac11000$ cannot be in $Y$ and that $999$ and $frac1999$ must be in $Y$. Any such $f$ and $Y$ which meets these conditions with $f(y)=frac1y$ for $y in Y$ will be a solution. So $f(500)$ can potentially take any value apart from $0$ or $1000$ or $frac11000$ or $500$
To construct an example, suppose you want $f(500)=k not in left0 , 1000,frac11000,500 right$. Then consider
$$
f(x) =
begincases
999 &quadtextif x= 1000 text or x= tfrac1999\
tfrac1999 &quadtextif x= 999\
tfrac1k &quadtextif x= k\
k &quadtextotherwise. \
endcases$$
There will be many other possible examples
answered Mar 20 at 17:27
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
Any $y$ in the image of $f$ will have $y=f(x)$ for some $x in mathbb R$ and then satisfy $y times f(y)=1$ so $f(y)=frac1y$ and thus $fleft(frac1yright)=y$. Let's call $f$'s image $Y=y in mathbb R mid exists x in mathbb R: y=f(x)$
You now know that $0$ cannot be in $Y$. Since $f(1000)=999$, you also know $1000$ and $frac11000$ cannot be in $Y$ and that $999$ and $frac1999$ must be in $Y$. Any such $f$ and $Y$ which meets these conditions with $f(y)=frac1y$ for $y in Y$ will be a solution. So $f(500)$ can potentially take any value apart from $0$ or $1000$ or $frac11000$ or $500$
To construct an example, suppose you want $f(500)=k not in left0 , 1000,frac11000,500 right$. Then consider
$$
f(x) =
begincases
999 &quadtextif x= 1000 text or x= tfrac1999\
tfrac1999 &quadtextif x= 999\
tfrac1k &quadtextif x= k\
k &quadtextotherwise. \
endcases$$
There will be many other possible examples
answered Mar 20 at 17:27
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
Any $y$ in the image of $f$ will have $y=f(x)$ for some $x in mathbb R$ and then satisfy $y times f(y)=1$ so $f(y)=frac1y$ and thus $fleft(frac1yright)=y$. Let's call $f$'s image $Y=y in mathbb R mid exists x in mathbb R: y=f(x)$
You now know that $0$ cannot be in $Y$. Since $f(1000)=999$, you also know $1000$ and $frac11000$ cannot be in $Y$ and that $999$ and $frac1999$ must be in $Y$. Any such $f$ and $Y$ which meets these conditions with $f(y)=frac1y$ for $y in Y$ will be a solution. So $f(500)$ can potentially take any value apart from $0$ or $1000$ or $frac11000$ or $500$
To construct an example, suppose you want $f(500)=k not in left0 , 1000,frac11000,500 right$. Then consider
$$
f(x) =
begincases
999 &quadtextif x= 1000 text or x= tfrac1999\
tfrac1999 &quadtextif x= 999\
tfrac1k &quadtextif x= k\
k &quadtextotherwise. \
endcases$$
There will be many other possible examples
answered Mar 20 at 17:27
HenryHenry
101k482169
$endgroup$
Any $y$ in the image of $f$ will have $y=f(x)$ for some $x in mathbb R$ and then satisfy $y times f(y)=1$ so $f(y)=frac1y$ and thus $fleft(frac1yright)=y$. Let's call $f$'s image $Y=y in mathbb R mid exists x in mathbb R: y=f(x)$
You now know that $0$ cannot be in $Y$. Since $f(1000)=999$, you also know $1000$ and $frac11000$ cannot be in $Y$ and that $999$ and $frac1999$ must be in $Y$. Any such $f$ and $Y$ which meets these conditions with $f(y)=frac1y$ for $y in Y$ will be a solution. So $f(500)$ can potentially take any value apart from $0$ or $1000$ or $frac11000$ or $500$
To construct an example, suppose you want $f(500)=k not in left0 , 1000,frac11000,500 right$. Then consider
$$
f(x) =
begincases
999 &quadtextif x= 1000 text or x= tfrac1999\
tfrac1999 &quadtextif x= 999\
tfrac1k &quadtextif x= k\
k &quadtextotherwise. \
endcases$$
There will be many other possible examples
answered Mar 20 at 17:27
HenryHenry
101k482169
answered Mar 20 at 17:27
HenryHenry
101k482169
answered Mar 20 at 17:27
HenryHenry
101k482169
answered Mar 20 at 17:27
HenryHenry
101k482169
101k482169
add a comment |
add a comment |
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