1. Problem A1: (a) Define the domain of a real-valued function $f(x)$. The domain is the set of all real numbers $x$ for which $f(x)$ is defined. (b) For $f(x) = 3x^2 + 2x - 1$, evaluate $f(a + h)$. Substitute $x = a + h$: $$f(a+h) = 3(a+h)^2 + 2(a+h) - 1$$ Expanding: $$= 3(a^2 + 2ah + h^2) + 2a + 2h - 1 = 3a^2 + 6ah + 3h^2 + 2a + 2h - 1$$ 2. Problem A2: (a) Find the inverse of $f(x) = 5 + \sqrt{2x} - 3 = 2 + \sqrt{2x}$. Set $y = 2 + \sqrt{2x}$, solve for $x$: $$y - 2 = \sqrt{2x}$$ Square both sides: $$(y - 2)^2 = 2x \implies x = \frac{(y - 2)^2}{2}$$ Therefore, $$f^{-1}(x) = \frac{(x - 2)^2}{2}$$ (b) Solve inequalities: (i) $5x + 1 > 7x - 7$ $$5x + 1 > 7x - 7 \implies 1 + 7 > 7x - 5x \implies 8 > 2x \implies x < 4$$ (ii) $x^2 - 6x \leq 16$ Bring all terms to one side: $$x^2 - 6x - 16 \leq 0$$ Factor quadratic: $$(x - 8)(x + 2) \leq 0$$ Using test intervals, solution is $$[-2, 8]$$ Intersection for both is $$[-2, 4)$$ (c) Function continuity at $x = 0$: Given $$f(x) = \frac{2x + \sin^{-1} x}{2x - \tan^{-1} x}, x \neq 0$$ $$f(0) = 2\varphi - 1$$ To be continuous: $$\lim_{x \to 0} f(x) = f(0)$$ Use Taylor expansions near zero: $\sin^{-1} x \approx x$ and $\tan^{-1} x \approx x$ So numerator: $$2x + x = 3x$$ Denominator: $$2x - x = x$$ Limit: $$\lim_{x \to 0} \frac{3x}{x} = 3$$ So, $$2\varphi - 1 = 3 \implies 2\varphi = 4 \implies \varphi = 2$$ 3. Problem A3: (a) Evaluate limits: (i) $$\lim_{x \to 8} (x^2 - 7x + 5) = 8^2 - 7 \times 8 + 5 = 64 - 56 + 5 = 13$$ (ii) $$\lim_{x \to 1}\frac{x^2 + x - 2}{x^2 - x}$$ Factor numerator and denominator: $$\frac{(x - 1)(x + 2)}{x(x - 1)}$$ Simplify by cancelling $(x-1)$ (for $x \neq 1$): $$\frac{x+2}{x}$$ Substitute $x=1$: $$\frac{1 + 2}{1} = 3$$ (b) For the limit $$\lim_{x \to 2} (x^2 + 3x) = 10$$, given $\varepsilon > 0$, find largest $\delta$ so: If $|x - 2| < \delta$ then $$|(x^2 + 3x) - 10| < \varepsilon$$ Simplify: $$|x^2 + 3x - 10| = |(x-2)(x+5)|$$ We want $$|(x-2)(x+5)| < \varepsilon$$ Choose neighborhood $|x-2| < 1$ so $x \in (1,3)$ and thus $$|x+5| < 8$$ Therefore, $$|x-2| < \frac{\varepsilon}{8}$$ So the largest $\delta$ that can be used is $$\delta = \min \{1, \frac{\varepsilon}{8} \}$$ Final answers summarized: A1: - Domain is all real numbers. - $f(a + h) = 3a^2 + 6ah + 3h^2 + 2a + 2h - 1$ A2: - $f^{-1}(x) = \frac{(x - 2)^2}{2}$ - $x \in [-2,4)$ - $\varphi = 2$ A3: - (i) 13 - (ii) 3 - $\delta = \min \{1, \frac{\varepsilon}{8} \}$