1. **Problem 1: Define and find values of the function** Given the function: $$f(n) = \begin{cases} 4 & \text{if } n=1 \text{ or } n=2 \\ 2f(n-1) + f(n-2) & \text{if } n \geq 3 \end{cases}$$ We need to find $f(1)$, $f(2)$, $f(3)$, $f(4)$, and $f(5)$. **Step 1:** Identify base cases: - $f(1) = 4$ - $f(2) = 4$ **Step 2:** Use the recursive formula for $n \geq 3$: $$f(n) = 2f(n-1) + f(n-2)$$ **Step 3:** Calculate $f(3)$: $$f(3) = 2f(2) + f(1) = 2 \times 4 + 4 = 8 + 4 = 12$$ **Step 4:** Calculate $f(4)$: $$f(4) = 2f(3) + f(2) = 2 \times 12 + 4 = 24 + 4 = 28$$ **Step 5:** Calculate $f(5)$: $$f(5) = 2f(4) + f(3) = 2 \times 28 + 12 = 56 + 12 = 68$$ **Final answers:** $$f(1) = 4, \quad f(2) = 4, \quad f(3) = 12, \quad f(4) = 28, \quad f(5) = 68$$ --- 2. **Problem 2: Domain and Range Questions** **a) Find the range of $f(x) = |x|$** **Step 1:** Recall that the absolute value function outputs the non-negative value of $x$. **Step 2:** Since $|x| \geq 0$ for all real $x$, the range is all real numbers greater than or equal to zero. **Answer:** $$\text{Range} = [0, \infty)$$ **b) Find the understood domain of the following functions:** **b1.** $f(x) = \sqrt{4 - x^2}$ **Step 1:** The radicand (expression inside the square root) must be non-negative: $$4 - x^2 \geq 0$$ **Step 2:** Solve inequality: $$x^2 \leq 4$$ $$-2 \leq x \leq 2$$ **Answer:** $$\text{Domain} = [-2, 2]$$ **b2.** $f(x) = \sqrt{\frac{x}{x^2 + x - 2}}$ **Step 1:** The radicand must be non-negative: $$\frac{x}{x^2 + x - 2} \geq 0$$ **Step 2:** Factor denominator: $$x^2 + x - 2 = (x + 2)(x - 1)$$ **Step 3:** Find critical points where numerator or denominator is zero: - Numerator zero at $x=0$ - Denominator zero at $x=-2$ and $x=1$ (excluded from domain) **Step 4:** Test intervals determined by these points: - $(-\infty, -2)$: Choose $x=-3$, numerator negative, denominator positive, fraction negative - $(-2, 0)$: Choose $x=-1$, numerator negative, denominator negative, fraction positive - $(0, 1)$: Choose $x=0.5$, numerator positive, denominator negative, fraction negative - $(1, \infty)$: Choose $x=2$, numerator positive, denominator positive, fraction positive **Step 5:** Domain includes intervals where fraction is non-negative and denominator not zero: $$(-2, 0] \cup (1, \infty)$$ --- 3. **Problem 3: Explain why $f: \mathbb{Z} \to \mathbb{N}$ defined by $f(x) = 2 - x^2$ is not a function** **Step 1:** Recall that a function must assign each input exactly one output in the codomain. **Step 2:** The codomain is $\mathbb{N}$ (natural numbers, usually positive integers). **Step 3:** For $x=0$, $$f(0) = 2 - 0^2 = 2 \in \mathbb{N}$$ **Step 4:** For $x=2$, $$f(2) = 2 - 4 = -2 \notin \mathbb{N}$$ **Step 5:** Since $f(2)$ is not in the codomain $\mathbb{N}$, the mapping is not a function from $\mathbb{Z}$ to $\mathbb{N}$. **Answer:** The function is not well-defined because some inputs map to values outside the codomain. ---