2. Solve the inequalities \( \begin{array}{ll}\text { (a) } \frac{1}{x+4} \leq 7 & \text { (b) } x+\frac{2}{x-1} \geq 4 \text {. } \\ \text { 3. Let } a, b \in[0, \infty) \text { with } a^{2} \leq b^{2} \text {. Show that } a \leq b \text {. }\end{array} \)

**2. (a) Solve the inequality \( \frac{1}{x+4} \leq 7 \).** We must take care of the denominator. First, note that \( x \neq -4 \) since the expression is undefined there. We consider two cases depending on the sign of \( x+4 \). *Case 1: \( x+4 > 0 \) (i.e. \( x > -4 \)).* In this case, we can multiply both sides of the inequality by \( x+4 \) without changing the inequality sign: \[ \frac{1}{x+4} \leq 7 \quad \Longrightarrow \quad 1 \leq 7(x+4). \] Expanding and solving for \( x \): \[ 1 \leq 7x + 28 \quad \Longrightarrow \quad 7x \geq 1 - 28 = -27 \quad \Longrightarrow \quad x \geq -\frac{27}{7}. \] Since we assumed \( x > -4 \) (or \( x > -\frac{28}{7} \)) and \(-\frac{27}{7}>-\frac{28}{7}\), the valid solutions in this case are: \[ x \geq -\frac{27}{7}. \] *Case 2: \( x+4 < 0 \) (i.e. \( x < -4 \)).* Here, multiplying both sides by \( x+4 \) (which is negative) reverses the inequality: \[ \frac{1}{x+4} \leq 7 \quad \Longrightarrow \quad 1 \geq 7(x+4). \] Expanding and solving: \[ 1 \geq 7x + 28 \quad \Longrightarrow \quad 7x \leq -27 \quad \Longrightarrow \quad x \leq -\frac{27}{7}. \] However, note that in this case we already have \( x < -4 \) (i.e. \( x < -\frac{28}{7} \)). Since \[ x < -\frac{28}{7} \implies x \leq -\frac{27}{7} \quad \text{(because } -\frac{28}{7} < -\frac{27}{7}\text{)}, \] the valid solutions in this region are simply: \[ x < -4. \] *Combining the cases:* The inequality holds for \[ x \in (-\infty, -4) \cup \left[-\frac{27}{7}, \infty\right), \] with the note that \( x \neq -4 \). --- **2. (b) Solve the inequality \( x+\frac{2}{x-1} \geq 4 \).** First, note that \( x\neq 1 \) because the denominator becomes zero. Rearrange the inequality by subtracting 4 from both sides: \[ x+\frac{2}{x-1} - 4 \geq 0. \] Combine terms over a common denominator: \[ \frac{(x-4)(x-1) + 2}{x-1} \geq 0. \] Expand the numerator: \[ (x-4)(x-1) = x^2 - 5x + 4, \] so that \[ x^2 - 5x + 4 + 2 = x^2 - 5x + 6. \] Factor the quadratic: \[ x^2 -5x+6 = (x-2)(x-3). \] The inequality becomes: \[ \frac{(x-2)(x-3)}{x-1} \geq 0. \] Identify the critical points where the expression is zero or undefined: \( x = 1 \) (undefined), \( x = 2 \), and \( x = 3 \). We now analyze the sign of the expression on the intervals determined by these points. 1. **For \( x < 1 \):** Choose \( x = 0 \). - Numerator: \( (0-2)(0-3) = (-2)(-3) = 6 \) (positive). - Denominator: \( 0-1 = -1 \) (negative). So, the expression is negative. 2. **For \( 1 < x < 2 \):** Choose \( x = 1.5 \). - Numerator: \( (1.5-2)(1.5-3) = (-0.5)(-1.5) = 0.75 \) (positive). - Denominator: \( 1.5-1 = 0.5 \) (positive). So, the expression is positive. 3. **For \( 2 < x < 3 \):** Choose \( x = 2.5 \). - Numerator: \( (2.5-2)(2.5-3) = (0.5)(-0.5) = -0.25 \) (negative). - Denominator: \( 2.5-1 = 1.5 \) (positive). So, the expression is negative. 4. **For \( x > 3 \):** Choose \( x = 4 \). - Numerator: \( (4-2)(4-3) = (2)(1) = 2 \) (positive). - Denominator: \( 4-1 = 3 \) (positive). So, the expression is positive. The expression is also zero when \( x = 2 \) or \( x = 3 \) (where the numerator is zero). Thus, the inequality \( \frac{(x-2)(x-3)}{x-1} \geq 0 \) holds for: \[ x \in (1,2] \cup [3,\infty), \] with the point \( x=1 \) excluded. --- **3. Let \( a, b \in [0, \infty) \) with \( a^2 \leq b^2 \). Show that \( a \leq b \).** Since \( a \) and \( b \) are nonnegative, we have: \[ \sqrt{a^2} = a \quad \text{and} \quad \sqrt{b^2} = b. \] The square root function on \([0, \infty)\) is an increasing function. Therefore, taking the square root on both sides of the inequality \( a^2 \leq b^2 \) preserves the inequality: \[ \sqrt{a^2} \leq \sqrt{b^2} \quad \Longrightarrow \quad a \leq b. \]