**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**

The **Answer to the Question**

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**Here's the Solution to this Question**

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.