1. **Problem Statement:** Determine which of the following statements are true for the functions $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to [0, \infty)$ defined by $f(x) = g(x) = x^2$ for every $x \in \mathbb{R}$: - (1) $f$ is one-to-one - (2) $f$ is onto - (3) $g$ is one-to-one - (4) $g$ is onto 2. **Definitions and Important Rules:** - A function is **one-to-one (injective)** if different inputs produce different outputs, i.e., $f(x_1) = f(x_2) \implies x_1 = x_2$. - A function is **onto (surjective)** if every element in the codomain has a preimage in the domain. 3. **Check if $f$ is one-to-one:** - Since $f(x) = x^2$, consider $f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$. - Here, $f(1) = f(-1)$ but $1 \neq -1$, so $f$ is **not one-to-one**. 4. **Check if $f$ is onto:** - The codomain of $f$ is $\mathbb{R}$ (all real numbers). - The function $f(x) = x^2$ only produces outputs $\geq 0$. - Negative numbers in $\mathbb{R}$ have no preimage, so $f$ is **not onto**. 5. **Check if $g$ is one-to-one:** - $g$ has the same formula as $f$, $g(x) = x^2$, but codomain is $[0, \infty)$. - Using the same example as for $f$, $g(1) = g(-1) = 1$ but $1 \neq -1$. - So $g$ is **not one-to-one**. 6. **Check if $g$ is onto:** - The codomain of $g$ is $[0, \infty)$. - For any $y \geq 0$, there exists $x = \sqrt{y}$ such that $g(x) = x^2 = y$. - Hence, $g$ is **onto**. **Final answers:** - (1) False - (2) False - (3) False - (4) True