Let's analyze each function for **injectivity** (one-to-one) and **surjectivity** (onto), including detailed explanations. --- ### (a) \( f : \mathbb{Z} \to \mathbb{Z} \) defined by \( f(x) = 3x + 1 \) #### **Injective?** Suppose \( f(x_1) = f(x_2) \): \[ 3x_1 + 1 = 3x_2 + 1 \implies 3x_1 = 3x_2 \implies x_1 = x_2 \] **Conclusion:** \( f \) is **injective**. #### **Surjective?** We need to see if for every \( y \in \mathbb{Z} \), there is an \( x \in \mathbb{Z} \) such that \( y = 3x + 1 \). \[ x = \frac{y-1}{3} \] But \( x \) is an integer only if \( y-1 \) is divisible by 3 (i.e., \( y \equiv 1 \pmod{3} \)). So not every integer is hit. **Conclusion:** \( f \) is **not surjective**. --- ### (b) \( F : \mathbb{Q} \to \mathbb{Q} \) defined by \( F(x) = 3x + 1 \) #### **Injective?** Same as above: \[ 3x_1 + 1 = 3x_2 + 1 \implies x_1 = x_2 \] **Conclusion:** \( F \) is **injective**. #### **Surjective?** Let \( y \in \mathbb{Q} \): \[ x = \frac{y-1}{3} \] For any rational \( y \), \( x \) is also rational. **Conclusion:** \( F \) is **surjective**. --- ### (c) \( g : \mathbb{R} \to \mathbb{R} \) defined by \( g(x) = x^3 \) #### **Injective?** \[ x_1^3 = x_2^3 \implies x_1 = x_2 \] **Conclusion:** \( g \) is **injective**. #### **Surjective?** For any \( y \in \mathbb{R} \), \( x = \sqrt[3]{y} \) is real. **Conclusion:** \( g \) is **surjective**. --- ### (d) \( G : \mathbb{Q} \to \mathbb{Q} \) defined by \( G(x) = x^3 \) #### **Injective?** If \( x_1^3 = x_2^3 \), then \( x_1 = x_2 \). **Conclusion:** \( G \) is **injective**. #### **Surjective?** For any \( y \in \mathbb{Q} \), \( x = \sqrt[3]{y} \) is rational **if and only if** \( y \) is a rational cube. But all rationals do have rational cube roots (cube roots of rationals are rational). Thus, for any \( y = \frac{a}{b} \in \mathbb{Q} \), \( x = \sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}} \) is rational if \( a, b \) are cubes of integers, but not in general for all rationals. For example, \( y = 2 \), \( x = \sqrt[3]{2} \notin \mathbb{Q} \). **Conclusion:** \( G \) is **injective but not surjective**. --- ### (e) \( k : \mathbb{R} \to \mathbb{R} \) defined by \( k(x) = e^{-x^2} \) #### **Injective?** Suppose \( k(x_1) = k(x_2) \): \[ e^{-x_1^2} = e^{-x_2^2} \implies -x_1^2 = -x_2^2 \implies x_1^2 = x_2^2 \implies x_1 = x_2 \text{ or } x_1 = -x_2 \] So not injective. #### **Surjective?** The range of \( k(x) \): - \( k(x) > 0 \) for all \( x \). - \( k(0) = 1 \), as \( |x| \to \infty \), \( k(x) \to 0 \). - So \( \text{Image}(k) = (0, 1] \), not all of \( \mathbb{R} \). **Conclusion:** \( k \) is **not injective** and **not surjective**. --- ## **Summary Table** | Function | Injective? | Surjective? | Justification | |----------|:----------:|:-----------:|--------------| | (a) \( f \) | Yes | No | Only maps to integers congruent to 1 mod 3 | | (b) \( F \) | Yes | Yes | All rationals are covered | | (c) \( g \) | Yes | Yes | All reals are covered, strictly increasing | | (d) \( G \) | Yes | No | Not all rationals are cubes of rationals | | (e) \( k \) | No | No | Not one-to-one, range is \( (0,1] \) | --- **If you need more detail for any part, let me know!**

Let's analyze each function step-by-step to determine whether it is injective (one-to-one) and surjective (onto). We'll include detailed calculations and reasoning for each. --- ### **(a) \(f : \mathbb{Z} \to \mathbb{Z}\) defined by \(f(x) = 3x + 1\)** #### **Injectivity** - **Definition:** \(f\) is injective if \(f(x_1) = f(x_2) \Rightarrow x_1 = x_2\). - **Calculation:** Suppose \(f(x_1) = f(x_2)\): \[ 3x_1 + 1 = 3x_2 + 1 \] Subtract 1 from both sides: \[ 3x_1 = 3x_2 \] Divide both sides by 3: \[ x_1 = x_2 \] - **Conclusion:** Since the only way for \(f(x_1) = f(x_2)\) is when \(x_1 = x_2\), **\(f\) is injective**. --- #### **Surjectivity** - **Definition:** \(f\) is surjective if for every \(y \in \mathbb{Z}\), there exists \(x \in \mathbb{Z}\) such that \(f(x) = y\). - **Calculation:** Given \(y \in \mathbb{Z}\), find \(x\): \[ y = 3x + 1 \Rightarrow x = \frac{y - 1}{3} \] For \(x\) to be an integer, \(\frac{y - 1}{3}\) must be an integer. - **Observation:** \(x\) is integer **only if** \(y - 1\) is divisible by 3. - **Counterexample:** Take \(y = 2\): \[ x = \frac{2 - 1}{3} = \frac{1}{3} \notin \mathbb{Z} \] Since \(y=2\) cannot be written as \(3x + 1\) with integer \(x\), \(f\) does **not** map onto all integers. - **Conclusion:** \(f\) is **not surjective**. --- ### **(b) \(F : \mathbb{Q} \to \mathbb{Q}\) defined by \(F(x) = 3x + 1\)** #### **Injectivity** - Same as above: \[ F(x_1) = F(x_2) \Rightarrow 3x_1 + 1 = 3x_2 + 1 \Rightarrow x_1 = x_2 \] - **Conclusion:** \(F\) is **injective**. #### **Surjectivity** - For any \(y \in \mathbb{Q}\): \[ x = \frac{y - 1}{3} \] - Since \(\frac{y - 1}{3}\) is rational for any rational \(y\), **every rational number** is hit. - **Conclusion:** \(F\) is **surjective**. --- ### **(c) \(g : \mathbb{R} \to \mathbb{R}\) defined by \(g(x) = x^3\)** #### **Injectivity** - **Calculation:** Suppose \(g(x_1) = g(x_2)\): \[ x_1^3 = x_2^3 \] - Cube function is strictly increasing over \(\mathbb{R}\): \[ x_1^3 = x_2^3 \Rightarrow x_1 = x_2 \] - **Conclusion:** \(g\) is **injective**. #### **Surjectivity** - For any \(y \in \mathbb{R}\), \[ x = \sqrt[3]{y} \] - The cube root is always real, so: \[ g(x) = y \] - **Conclusion:** \(g\) is **surjective**. --- ### **(d) \(G : \mathbb{Q} \to \mathbb{Q}\) defined by \(G(x) = x^3\)** #### **Injectivity** - Suppose \(G(x_1) = G(x_2)\): \[ x_1^3 = x_2^3 \Rightarrow x_1 = x_2 \] - **Conclusion:** \(G\) is **injective**. #### **Surjectivity** - For any \(y \in \mathbb{Q}\), the pre-image: \[ x = \sqrt[3]{y} \] - **Question:** Is \(\sqrt[3]{y}\) always rational? - **Answer:** Not necessarily. For example, take \(y=2\): \[ x = \sqrt[3]{2} \notin \mathbb{Q} \] since 2 is not a perfect cube rational number. - **Conclusion:** \(G\) is **not surjective** because not all rationals are cubes of rationals. --- ### **(e) \(k : \mathbb{R} \to \mathbb{R}\) defined by \(k(x) = e^{-x^2}\)** #### **Injectivity** - **Observation:** \[ k(x) = e^{-x^2} \] - Since \(x^2 = (-x)^2\), for each \(x\), \[ k(x) = k(-x) \] - **Implication:** \(k\) is **not injective** because \(k(x) = k(-x)\) for all \(x\), and \(x \neq -x\) unless \(x=0\). - **Counterexample:** \[ k(1) = e^{-1} = k(-1) \] - **Conclusion:** \(k\) is **not injective**. #### **Surjectivity** - Range of \(k(x)\): \[ k(x) = e^{-x^2} \in (0, 1] \] - Because: \[ e^{-x^2} > 0 \quad \text{and} \quad e^{0} = 1 \] - As \(|x| \to \infty\), \(k(x) \to 0\). - **Range:** \((0, 1]\). It does **not** cover negative real numbers or numbers greater than 1. - **Conclusion:** \(k\) is **not surjective** onto \(\mathbb{R}\). --- ## **Final Summary:** | Function | **Injective?** | **Surjective?** | **Explanation** | |------------|:--------------:|:--------------:|-----------------| | (a) \(f(x) = 3x + 1\) | **Yes** | **No** | Not all integers (only those congruent to 1 mod 3). | | (b) \(F(x) = 3x + 1\) | **Yes** | **Yes** | All rationals covered. | | (c) \(g(x) = x^3\) | **Yes** | **Yes** | Monotonically increasing, covers all reals. | | (d) \(G(x) = x^3\) | **Yes** | **No** | Not all rationals are cubes of rationals. | | (e) \(k(x) = e^{-x^2}\) | **No** | **No** | Symmetric about 0, range only in \((0,1]\). | If you have further questions or need clarifications, feel free to ask!

Certainly! Here are the final answers for each part: **(a)** The function \(f: \mathbb{Z} \to \mathbb{Z}\) defined by \(f(x) = 3x + 1\) is **injective** but **not surjective**. **(b)** The function \(F: \mathbb{Q} \to \mathbb{Q}\) defined by \(F(x) = 3x + 1\) is **both injective and surjective** (bijective). **(c)** The function \(g: \mathbb{R} \to \mathbb{R}\) defined by \(g(x) = x^3\) is **both injective and surjective** (bijective). **(d)** The function \(G: \mathbb{Q} \to \mathbb{Q}\) defined by \(G(x) = x^3\) is **injective** but **not surjective**. **(e)** The function \(k: \mathbb{R} \to \mathbb{R}\) defined by \(k(x) = e^{-x^2}\) is **neither injective nor surjective**.