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Archangel Macsika

Determine whether each of these functions is a bijection from R to R (real).a) f (x) = 2x + 1b) f (x) = x2 + 1c) f (x) = x3

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Let us determine whether each of these functions is a bijection from \R to \R.


a) f (x) = 2x + 1


Let f(x_1)=f(x_2). Then 2x_1+1=2x_2+1, and thus 2x_1=2x_2. We conclude that x_1=x_2, and hence the function f is one-to-one.

For any y\in\R there exists x=\frac{y-1}2\in\R such that f(x)=f(\frac{y-1}2)=2\frac{y-1}2+1=y-1+1=y,

and hence the function f is a surjection.

Consequently, the function f is a bijection.


b) f (x) = x^2 + 1


Since f (-1) = (-1)^2 + 1=1^2+1=f(1), we conclude that this function is not one-to-one, and therefore f is not a bijection.


c) f (x) = x^3


Let f(x_1)=f(x_2). Then x_1^3=x_2^3, and thus x_1=x_2. We conclude that the function f is one-to-one.

For any y\in\R there exists x=\sqrt[3]{y}\in\R such that f(x)=f(\sqrt[3]{y})=(\sqrt[3]{y})^3=y,

and hence the function f is a surjection.

Consequently, the function f is a bijection.

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