**a) Consider the following functional relation, f, defined as: f : R \rightarrow R, f(x) = x2 Determine whether or not f is a bijection. If it is, prove it. If it is not, show why it is not. b) Consider the set F = {y | y = ax3 + b}, a, b being constants such that a \ne 0 and x \in R. Is F equivalent to R? If so, prove it. If not, explain in details why it is not the case.**

The **Answer to the Question**

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

a.)

$f:\R \to \R, f(x)=x^2$

This function is not a bijection since it is not one-to-one. We can see this as follows:

$f(-2)=f(2)=4$, but $-2 \neq 2$. Thus $f$ is not injective. Hence it is not a bijection.

b.)

$F=\{y|y=ax^3+b\}$

Yes, $F$ is equivalent to $\R$ since will can always find a bijection between the two set.

$f:F\to \R$