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)

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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 $(-1,1)$?







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  • 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















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 $(-1,1)$?







share|cite|improve this question





















  • 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













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 $(-1,1)$?







share|cite|improve this question













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 $(-1,1)$?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








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

















  • 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
















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