1. **Statement of the problem:** We have two exercises involving functions and polynomials. --- ### Exercise 1 Given the function $$f(x) = \frac{2x - 1}{x - 1}$$ defined on $$\mathbb{R} \setminus \{1\}$$. 1) Show that $$f(x) = 2 + \frac{1}{x - 1}$$ for all $$x \neq 1$$. 2) Study the monotonicity of $$f$$ on $$]-\infty, 1[$$ and $$]1, +\infty[$$. 3) Explain how to sketch the graph of $$f$$ and then draw it. 4) Prove that the point $$(1, 2)$$ is the center of symmetry of the curve. 5) Draw the graph of $$g(x) = |f(x)|$$ on the same coordinate system. 6) Consider the function $$h$$ defined on $$]1, +\infty[ \cup ]-\infty, \frac{1}{2}]$$ by $$h(x) = \sqrt{\frac{2x - 1}{x - 1}}$$. - Show that $$h$$ is a composition of $$f$$ and another function. - Deduce the monotonicity of $$h$$ on $$]\frac{1}{2}, +\infty[$$ and $$]-\infty, 1[$$. --- ### Exercise 2 Given the polynomial $$f(x) = x^3 - 6x^2 + 11x - 6$$. - Calculate $$f(0)$$ and $$f(3)$$ and deduce something. - Find real numbers $$\alpha, \beta, \delta$$ such that $$f(x) = (x - 3)(\alpha x^2 + \beta x + \delta)$$. - Solve in $$\mathbb{R}$$ the equation $$x^3 - 3x + 2 = 0$$. - Deduce the solutions of $$f(x) = 0$$. - Solve the inequality $$f(x) < 0$$ in $$\mathbb{R}$$. --- ### Solutions **Exercise 1:** 1. Start with the given function: $$f(x) = \frac{2x - 1}{x - 1}$$ Rewrite the numerator: $$2x - 1 = 2(x - 1) + 1$$ So, $$f(x) = \frac{2(x - 1) + 1}{x - 1} = \frac{2(x - 1)}{x - 1} + \frac{1}{x - 1} = 2 + \frac{1}{x - 1}$$ This proves the first part. 2. To study monotonicity, compute the derivative: $$f'(x) = \frac{d}{dx} \left(2 + \frac{1}{x - 1}\right) = -\frac{1}{(x - 1)^2}$$ Since $$ (x - 1)^2 > 0 $$ for all $$x \neq 1$$, we have: $$f'(x) < 0$$ for all $$x \neq 1$$. Therefore, $$f$$ is strictly decreasing on both intervals $$]-\infty, 1[$$ and $$]1, +\infty[$$. 3. To sketch the graph: - The function has a vertical asymptote at $$x = 1$$ because the denominator is zero. - The horizontal asymptote is $$y = 2$$ since $$f(x) \to 2$$ as $$x \to \pm \infty$$. - The function decreases on both sides of $$x=1$$. - Plot points and asymptotes, then draw the curve accordingly. 4. To prove $$(1, 2)$$ is a center of symmetry: Check if for all $$x$$, $$f(2 - x) + f(x) = 4$$ Calculate: $$f(2 - x) = 2 + \frac{1}{(2 - x) - 1} = 2 + \frac{1}{1 - x} = 2 - \frac{1}{x - 1}$$ Add to $$f(x)$$: $$f(x) + f(2 - x) = \left(2 + \frac{1}{x - 1}\right) + \left(2 - \frac{1}{x - 1}\right) = 4$$ This symmetry relation shows $$(1, 2)$$ is the center of symmetry. 5. The graph of $$g(x) = |f(x)|$$ is the absolute value of $$f(x)$$. - Reflect the parts of $$f(x)$$ below the x-axis above the x-axis. - Draw accordingly on the same axes. 6. For $$h(x) = \sqrt{\frac{2x - 1}{x - 1}}$$: Note that: $$f(x) = 2 + \frac{1}{x - 1} = \frac{2x - 1}{x - 1}$$ So, $$h(x) = \sqrt{f(x)}$$ Hence, $$h = \sqrt{\cdot} \circ f$$, composition of $$f$$ and the square root function. Domain of $$h$$ requires $$f(x) \geq 0$$. Study monotonicity: - Since $$f$$ is decreasing and square root is increasing on $$[0, +\infty[$$, the monotonicity of $$h$$ follows the sign and domain of $$f$$. - On $$]1/2, +\infty[$$, $$f(x) > 0$$ and $$h$$ is decreasing. - On $$]-\infty, 1/2]$$, $$h$$ is not defined because the radicand is negative or zero. --- **Exercise 2:** 1. Calculate: $$f(0) = 0 - 0 + 0 - 6 = -6$$ $$f(3) = 27 - 54 + 33 - 6 = 0$$ Since $$f(3) = 0$$, $$x=3$$ is a root. 2. Factorize: $$f(x) = (x - 3)(\alpha x^2 + \beta x + \delta)$$ Expand: $$(x - 3)(\alpha x^2 + \beta x + \delta) = \alpha x^3 + \beta x^2 + \delta x - 3\alpha x^2 - 3\beta x - 3\delta$$ Group terms: $$= \alpha x^3 + (\beta - 3\alpha) x^2 + (\delta - 3\beta) x - 3\delta$$ Match coefficients with $$f(x) = x^3 - 6x^2 + 11x - 6$$: - $$\alpha = 1$$ - $$\beta - 3 = -6 \Rightarrow \beta = -3$$ - $$\delta - 3\beta = 11 \Rightarrow \delta - 3(-3) = 11 \Rightarrow \delta + 9 = 11 \Rightarrow \delta = 2$$ - $$-3\delta = -6 \Rightarrow \delta = 2$$ (consistent) So, $$f(x) = (x - 3)(x^2 - 3x + 2)$$ 3. Solve $$x^3 - 3x + 2 = 0$$. Try rational roots: factors of 2 are $$\pm1, \pm2$$. - $$x=1: 1 - 3 + 2 = 0$$ root. - $$x=-2: -8 + 6 + 2 = 0$$ root. Divide polynomial by $$(x - 1)(x + 2) = x^2 + x - 2$$: Divide $$x^3 - 3x + 2$$ by $$x^2 + x - 2$$: Quotient is $$x - 1$$. So, $$x^3 - 3x + 2 = (x - 1)^2 (x + 2)$$ Roots are $$x=1$$ (double root) and $$x=-2$$. 4. Solve $$f(x) = 0$$: Recall: $$f(x) = (x - 3)(x^2 - 3x + 2) = (x - 3)(x - 1)(x - 2)$$ Roots are $$x=3, 1, 2$$. 5. Solve $$f(x) < 0$$: Sign analysis of factors: - $$x - 3$$ changes sign at 3 - $$x - 1$$ changes sign at 1 - $$x - 2$$ changes sign at 2 Intervals: - $$]-\infty, 1[$$: all factors negative or positive? At $$x=0$$: $$(0-3)(0-1)(0-2) = (-3)(-1)(-2) = -6 < 0$$ So $$f(x) < 0$$ on $$]-\infty, 1[$$. - $$]1, 2[$: At $$x=1.5$$: $$(1.5-3)(1.5-1)(1.5-2) = (-1.5)(0.5)(-0.5) = 0.375 > 0$$ - $$]2, 3[$: At $$x=2.5$$: $$(2.5-3)(2.5-1)(2.5-2) = (-0.5)(1.5)(0.5) = -0.375 < 0$$ - $$]3, +\infty[$: At $$x=4$$: $$(4-3)(4-1)(4-2) = (1)(3)(2) = 6 > 0$$ So $$f(x) < 0$$ on $$]-\infty, 1[ \cup ]2, 3[$. --- **Final answers:** - Exercise 1: Verified formula, monotonicity decreasing on both intervals, center of symmetry at $$(1, 2)$$, $$g(x) = |f(x)|$$ graph is absolute value, $$h$$ is composition $$h = \sqrt{\cdot} \circ f$$. - Exercise 2: $$f(0) = -6$$, $$f(3) = 0$$, factorization $$f(x) = (x - 3)(x^2 - 3x + 2)$$, roots of $$x^3 - 3x + 2 = 0$$ are $$1$$ (double root) and $$-2$$, roots of $$f(x) = 0$$ are $$1, 2, 3$$, and $$f(x) < 0$$ on $$]-\infty, 1[ \cup ]2, 3[$.