1. **Problem 01**: Given the function $f(x) = 3x^2 - x + 3$, find the following values: (a) $f(2)$ Calculate by substituting $x=2$: $$f(2) = 3(2)^2 - 2 + 3 = 3 \times 4 - 2 + 3 = 12 - 2 + 3 = 13$$ (b) $f(-2)$ Substitute $x = -2$: $$f(-2) = 3(-2)^2 - (-2) + 3 = 3 \times 4 + 2 + 3 = 12 + 2 + 3 = 17$$ (c) $f(t)$ Substitute $x = t$ (variable): $$f(t) = 3t^2 - t + 3$$ (d) $f(2x - 1)$ Replace $x$ by $2x - 1$: $$f(2x-1) = 3(2x-1)^2 - (2x-1) + 3$$ Expand $(2x-1)^2 = 4x^2 - 4x + 1$: $$= 3(4x^2 - 4x + 1) - 2x + 1 + 3 = 12x^2 - 12x + 3 - 2x + 1 + 3$$ Combine like terms: $$= 12x^2 - 14x + 7$$ (e) Compute $f(a^2)$ and $[f(a)]^2$: $f(a^2)$: $$= 3(a^2)^2 - a^2 + 3 = 3a^4 - a^2 + 3$$ $[f(a)]^2$: First find $f(a)$: $$f(a) = 3a^2 - a + 3$$ Square it: $$[f(a)]^2 = (3a^2 - a + 3)^2$$ Expand using $(A-B+C)^2 = A^2 - 2AB + 2AC + B^2 - 2BC + C^2$ method or FOIL: $$= (3a^2)^2 - 2 \times 3a^2 \times a + 2 \times 3a^2 \times 3 + (-a)^2 - 2 \times (-a) \times 3 + 3^2$$ Calculate: $$= 9a^4 - 6a^3 + 18a^2 + a^2 + 6a + 9 = 9a^4 - 6a^3 + 19a^2 + 6a + 9$$ (f) Find the expression $$\frac{f(x)-f(1)}{x-1}$$ Calculate $f(1)$: $$f(1) = 3(1)^2 - 1 + 3 = 3 - 1 + 3 = 5$$ Substitute in the numerator: $$\frac{(3x^2 - x + 3) - 5}{x-1} = \frac{3x^2 - x + 3 - 5}{x-1} = \frac{3x^2 - x - 2}{x-1}$$ Factor numerator: $$3x^2 - x - 2 = (3x + 2)(x - 1)$$ Simplify: $$\frac{(3x + 2)(x-1)}{x-1} = 3x + 2\quad \text{for } x \neq 1$$ 2. **Problem 02**: Find domain of each function: (a) $$f(x) = \frac{5x + 4}{x^2 + 3x + 2}$$ The denominator must be nonzero: $$x^2 + 3x + 2 \neq 0$$ Factor denominator: $$(x + 1)(x + 2) \neq 0$$ Exclude $x = -1$ and $x = -2$, so: $$\text{Domain} = \{x \in \mathbb{R} : x \neq -1, x \neq -2\}$$ (b) $$G(u) = \sqrt{u} + \sqrt{4 - u}$$ Radicands must be nonnegative: $$u \geq 0$$ $$4 - u \geq 0 \Rightarrow u \leq 4$$ Combine: $$0 \leq u \leq 4$$ (c) $$f(x) = \sqrt{5 - 5x - x^2}$$ Radicand must be nonnegative: $$5 - 5x - x^2 \geq 0$$ Rewrite: $$-x^2 - 5x + 5 \geq 0$$ Multiply both sides by $-1$ (reverse inequality): $$x^2 + 5x - 5 \leq 0$$ Find roots using quadratic formula: $$x = \frac{-5 \pm \sqrt{25 + 20}}{2} = \frac{-5 \pm \sqrt{45}}{2} = \frac{-5 \pm 3\sqrt{5}}{2}$$ The parabola opens upwards, so inequality holds between roots: $$\frac{-5 - 3\sqrt{5}}{2} \leq x \leq \frac{-5 + 3\sqrt{5}}{2}$$ 3. **Problem 03**: Based on the described graph: (a) State $f(-1)$: Graph starts above $y=1$ on left, so estimate $f(-1) \approx$ value $>1$ (roughly 2). (b) Estimate $f(2)$: Graph slightly above $y=1$ at $x=1$ and levels off, so $f(2) \approx 1.5$ (approximate). (c) Values of $x$ where $f(x) = 2$: From graph, $f(x)$ is 2 near $x = -1$, and possibly another value after $1$, say $x \approx 1.5$. (d) Values where $f(x) = 0$: Graph crosses $y=0$ near $x=0$, so $f(0) = 0$, estimate roots near $x=0$. (e) Domain and Range: Domain is all real numbers (graph stretches left and right). Range includes values from minimum below zero to above 2, that is $$\text{Range} = [\text{minimum} < 0, \text{maximum} > 2]$$ Final summarized answers are above with all steps shown.