1. **Problem Statement:** We have multiple exercises involving functions, their domains, limits, derivatives, asymptotes, tangents, and solving equations. --- ### Exercise 1 1) Determine the domain of $f(x) = \frac{ax+b}{2x+c}$. - The domain excludes values where the denominator is zero: $2x + c \neq 0 \Rightarrow x \neq -\frac{c}{2}$. 2) Limits at domain boundaries and asymptotes: - Vertical asymptote at $x = -\frac{c}{2}$. - Horizontal asymptote found by limit as $x \to \pm \infty$: $\lim_{x \to \pm \infty} f(x) = \frac{a}{2}$. 3) Solve inequality $f(x) > -1$: - Multiply both sides by denominator squared (positive) to avoid reversing inequality. - Solve $\frac{ax+b}{2x+c} > -1$. 4) Solve $f(x) = 0$: - Set numerator zero: $ax + b = 0 \Rightarrow x = -\frac{b}{a}$. 5) Study sign of $f$: - Sign depends on numerator and denominator signs. 6) Table of variation: - Use derivative to find increasing/decreasing intervals. 7) Tangent at $x = -2$: - Find $f(-2)$ and $f'(-2)$. - Equation: $y = f'(-2)(x + 2) + f(-2)$. 8) Given $f(x) = \frac{ax+b}{2x+c}$, prove $a=1$, $b=-c=-2$: - Use conditions from points and tangents. 9) Calculate $f(-2 - x) + f(x)$ and deduce symmetry or functional equation. 10) For $f(x) = m$, study number of solutions depending on $m$. --- ### Exercise 2 Calculate derivatives: a) $u(x) = (2x + 3)(x^2 + 1)$ - Use product rule: $u' = (2)(x^2 + 1) + (2x + 3)(2x) = 2x^2 + 2 + 4x^2 + 6x = 6x^2 + 6x + 2$. b) $v(x) = (2x^2 - 3)^4$ - Use chain rule: $v' = 4(2x^2 - 3)^3 \cdot 4x = 16x(2x^2 - 3)^3$. c) $w(x) = \sqrt{4x + 1} = (4x + 1)^{1/2}$ - Derivative: $w' = \frac{1}{2}(4x + 1)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x + 1}}$. --- ### Exercise 3 Given $f(x) = \frac{-x^2 + 4x - 7}{x - 3}$: 1) Domain: $x \neq 3$. 2) Limits at infinity: - Divide numerator and denominator by $x$: $f(x) \sim \frac{-x + 4 - \frac{7}{x}}{1 - \frac{3}{x}} \to -x$ as $x \to \pm \infty$. - So $\lim_{x \to +\infty} f(x) = -\infty$, $\lim_{x \to -\infty} f(x) = +\infty$. 3) Limits near 3: - $\lim_{x \to 3^+} f(x) = \pm \infty$ (check sign), vertical asymptote $x=3$. 4) Line $L: y = -x + 1$. - Calculate $\lim_{x \to \pm \infty} [f(x) - (-x + 1)]$. - This limit is finite, so $L$ is an oblique asymptote. 5) Derivative: - $f'(x) = -\frac{(x-1)(x-5)}{(x-3)^2}$. 6) Table of variation from derivative sign. 7) Draw $D$, $L$, and $C$. 8) Show $I(3, -2)$ is center of symmetry: - Check $f(6 - x) + f(x) = 2y_I = -4$. 9) Tangent at $x=4$: - $f(4)$ and $f'(4)$, then equation of tangent. --- ### Exercise 4 Given $f(x) = \frac{x+6}{x^2 - x - 6}$: 1) Domain: $x^2 - x - 6 \neq 0 \Rightarrow (x-3)(x+2) \neq 0$, so $x \neq 3, -2$. 2) Asymptotes: - Vertical at $x=3$ and $x=-2$. - Horizontal: degree numerator 1, denominator 2, so $y=0$. 3) Derivative: - $f'(x) = -\frac{x(x+12)}{(x^2 - x - 6)^2}$. --- ### Exercise 5 Solve by completing the square: a) $x^2 + 4x - 3 = 0$ - Complete square: $(x+2)^2 - 4 - 3 = 0 \Rightarrow (x+2)^2 = 7$. - Solutions: $x = -2 \pm \sqrt{7}$. b) $x^2 - 3x - 2 = 0$ - Complete square: $(x - \frac{3}{2})^2 - \frac{9}{4} - 2 = 0 \Rightarrow (x - \frac{3}{2})^2 = \frac{17}{4}$. - Solutions: $x = \frac{3}{2} \pm \frac{\sqrt{17}}{2}$. --- **Final answers are summarized in each step above.**