Consider the function $f:mathbb R to mathbb R$ defined by $f(x)=(x+2)^3$. The set $f^−1((−1,1))$ is? (For Improvement)
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This question may be related to determining one to one functions under algebric tests.
1) A function $f$ is said to be one-one or (injective) if;
$f(x_1) = f(x_2)$ which implies $x_1 = x_2$
2) Lemma 2. The function $f$ is one-to-one if and only if
$forall x_1, forall x_2, x_1$ is not equal to $x_2$ implies $f(x_1)$ is not equal to $f(x_2)$.
Would I be right to state the set is $(-1,1)$?
algebra-precalculus functions
 |Â
show 1 more comment
up vote
0
down vote
favorite
This question may be related to determining one to one functions under algebric tests.
1) A function $f$ is said to be one-one or (injective) if;
$f(x_1) = f(x_2)$ which implies $x_1 = x_2$
2) Lemma 2. The function $f$ is one-to-one if and only if
$forall x_1, forall x_2, x_1$ is not equal to $x_2$ implies $f(x_1)$ is not equal to $f(x_2)$.
Would I be right to state the set is $(-1,1)$?
algebra-precalculus functions
Yes both for the stated just failed to type them well. Both are powers
– Lamech Muluya Kibudde
3 hours ago
The function is one to one and onto, hence has an inverse, can you find the inverse function?
– kingW3
3 hours ago
I asks what set is the image of the set $(-1,1)$ under $f^-1$.
– mvw
3 hours ago
Please look at the way I reformatted the title to see if that is what you intended.
– saulspatz
2 hours ago
$f^-1((-1,1))neq(-1,1).$ Note that $f^-1(0) = -2notin(-1,1)$
– saulspatz
2 hours ago
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question may be related to determining one to one functions under algebric tests.
1) A function $f$ is said to be one-one or (injective) if;
$f(x_1) = f(x_2)$ which implies $x_1 = x_2$
2) Lemma 2. The function $f$ is one-to-one if and only if
$forall x_1, forall x_2, x_1$ is not equal to $x_2$ implies $f(x_1)$ is not equal to $f(x_2)$.
Would I be right to state the set is $(-1,1)$?
algebra-precalculus functions
This question may be related to determining one to one functions under algebric tests.
1) A function $f$ is said to be one-one or (injective) if;
$f(x_1) = f(x_2)$ which implies $x_1 = x_2$
2) Lemma 2. The function $f$ is one-to-one if and only if
$forall x_1, forall x_2, x_1$ is not equal to $x_2$ implies $f(x_1)$ is not equal to $f(x_2)$.
Would I be right to state the set is $(-1,1)$?
algebra-precalculus functions
edited 2 hours ago


saulspatz
10.2k21323
10.2k21323
asked 3 hours ago
Lamech Muluya Kibudde
113
113
Yes both for the stated just failed to type them well. Both are powers
– Lamech Muluya Kibudde
3 hours ago
The function is one to one and onto, hence has an inverse, can you find the inverse function?
– kingW3
3 hours ago
I asks what set is the image of the set $(-1,1)$ under $f^-1$.
– mvw
3 hours ago
Please look at the way I reformatted the title to see if that is what you intended.
– saulspatz
2 hours ago
$f^-1((-1,1))neq(-1,1).$ Note that $f^-1(0) = -2notin(-1,1)$
– saulspatz
2 hours ago
 |Â
show 1 more comment
Yes both for the stated just failed to type them well. Both are powers
– Lamech Muluya Kibudde
3 hours ago
The function is one to one and onto, hence has an inverse, can you find the inverse function?
– kingW3
3 hours ago
I asks what set is the image of the set $(-1,1)$ under $f^-1$.
– mvw
3 hours ago
Please look at the way I reformatted the title to see if that is what you intended.
– saulspatz
2 hours ago
$f^-1((-1,1))neq(-1,1).$ Note that $f^-1(0) = -2notin(-1,1)$
– saulspatz
2 hours ago
Yes both for the stated just failed to type them well. Both are powers
– Lamech Muluya Kibudde
3 hours ago
Yes both for the stated just failed to type them well. Both are powers
– Lamech Muluya Kibudde
3 hours ago
The function is one to one and onto, hence has an inverse, can you find the inverse function?
– kingW3
3 hours ago
The function is one to one and onto, hence has an inverse, can you find the inverse function?
– kingW3
3 hours ago
I asks what set is the image of the set $(-1,1)$ under $f^-1$.
– mvw
3 hours ago
I asks what set is the image of the set $(-1,1)$ under $f^-1$.
– mvw
3 hours ago
Please look at the way I reformatted the title to see if that is what you intended.
– saulspatz
2 hours ago
Please look at the way I reformatted the title to see if that is what you intended.
– saulspatz
2 hours ago
$f^-1((-1,1))neq(-1,1).$ Note that $f^-1(0) = -2notin(-1,1)$
– saulspatz
2 hours ago
$f^-1((-1,1))neq(-1,1).$ Note that $f^-1(0) = -2notin(-1,1)$
– saulspatz
2 hours ago
 |Â
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873299%2fconsider-the-function-f-mathbb-r-to-mathbb-r-defined-by-fx-x23-the%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Yes both for the stated just failed to type them well. Both are powers
– Lamech Muluya Kibudde
3 hours ago
The function is one to one and onto, hence has an inverse, can you find the inverse function?
– kingW3
3 hours ago
I asks what set is the image of the set $(-1,1)$ under $f^-1$.
– mvw
3 hours ago
Please look at the way I reformatted the title to see if that is what you intended.
– saulspatz
2 hours ago
$f^-1((-1,1))neq(-1,1).$ Note that $f^-1(0) = -2notin(-1,1)$
– saulspatz
2 hours ago