Does the image of $f(x)+f(f(x))$ having a hole imply that the image of $f(x)$ has a hole?
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Let $f:mathbb{R} to mathbb{R}$ be an injective function. Assume there exists $a$ such that $forall x in mathbb{R}, f(x)+f(f(x)) neq a$. Does that imply that $exists b, forall x in mathbb{R}, f(x) neq b$?
This question is a lemma that would be helpful in my attempt to answer this question. I have a feeling that this is false, but have been unable to construct a counterexample.
real-analysis algebra-precalculus functions
asked Nov 12 at 23:49
DreamConspiracy
8261216
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up vote
1
down vote
favorite
Let $f:mathbb{R} to mathbb{R}$ be an injective function. Assume there exists $a$ such that $forall x in mathbb{R}, f(x)+f(f(x)) neq a$. Does that imply that $exists b, forall x in mathbb{R}, f(x) neq b$?
This question is a lemma that would be helpful in my attempt to answer this question. I have a feeling that this is false, but have been unable to construct a counterexample.
real-analysis algebra-precalculus functions
asked Nov 12 at 23:49
DreamConspiracy
8261216
2
What about $f(x)=-x$?
– Clayton
Nov 13 at 0:08
@Clayton I can't believe I didn't see that. Thank you
– DreamConspiracy
Nov 13 at 0:09
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f:mathbb{R} to mathbb{R}$ be an injective function. Assume there exists $a$ such that $forall x in mathbb{R}, f(x)+f(f(x)) neq a$. Does that imply that $exists b, forall x in mathbb{R}, f(x) neq b$?
This question is a lemma that would be helpful in my attempt to answer this question. I have a feeling that this is false, but have been unable to construct a counterexample.
real-analysis algebra-precalculus functions
asked Nov 12 at 23:49
DreamConspiracy
8261216
Let $f:mathbb{R} to mathbb{R}$ be an injective function. Assume there exists $a$ such that $forall x in mathbb{R}, f(x)+f(f(x)) neq a$. Does that imply that $exists b, forall x in mathbb{R}, f(x) neq b$?
This question is a lemma that would be helpful in my attempt to answer this question. I have a feeling that this is false, but have been unable to construct a counterexample.
real-analysis algebra-precalculus functions
real-analysis algebra-precalculus functions
asked Nov 12 at 23:49
DreamConspiracy
8261216
asked Nov 12 at 23:49
DreamConspiracy
8261216
asked Nov 12 at 23:49
DreamConspiracy
8261216
asked Nov 12 at 23:49
DreamConspiracy
8261216
asked Nov 12 at 23:49
DreamConspiracy
8261216
8261216
2
What about $f(x)=-x$?
– Clayton
Nov 13 at 0:08
@Clayton I can't believe I didn't see that. Thank you
– DreamConspiracy
Nov 13 at 0:09
add a comment |
2
What about $f(x)=-x$?
– Clayton
Nov 13 at 0:08
@Clayton I can't believe I didn't see that. Thank you
– DreamConspiracy
Nov 13 at 0:09
2
2
What about $f(x)=-x$?
– Clayton
Nov 13 at 0:08
What about $f(x)=-x$?
– Clayton
Nov 13 at 0:08
@Clayton I can't believe I didn't see that. Thank you
– DreamConspiracy
Nov 13 at 0:09
@Clayton I can't believe I didn't see that. Thank you
– DreamConspiracy
Nov 13 at 0:09
add a comment |
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2
What about $f(x)=-x$?
– Clayton
Nov 13 at 0:08
@Clayton I can't believe I didn't see that. Thank you
– DreamConspiracy
Nov 13 at 0:09