For the function \( h(x)=\frac{3 x}{(x-7)(x+8)} \), solve the following inequality. \( h(x) \geq 0 \)

1. Write the function: \[ h(x)=\frac{3x}{(x-7)(x+8)} \] 2. Identify the critical points where either the numerator or denominator is zero: - Numerator zero: \[ 3x=0\quad \Longrightarrow \quad x=0. \] - Denominator zero: \[ x-7=0\quad \Longrightarrow \quad x=7; \] \[ x+8=0\quad \Longrightarrow \quad x=-8. \] The points \( x=-8 \) and \( x=7 \) are not in the domain. 3. Divide the real line into intervals determined by the points \( -8 \), \( 0 \), and \( 7 \): - \( (-\infty,-8) \) - \( (-8,0) \) - \( (0,7) \) - \( (7,\infty) \) 4. Choose a test point in each interval to determine the sign of \( h(x) \). - For \( x \in (-\infty,-8) \) (e.g., \( x=-10 \)): - Numerator: \( 3(-10)=-30 \) (negative). - Denominator: \[ (-10-7)(-10+8)=(-17)(-2)=34 \quad \text{(positive)}. \] - So, \( h(x) \) is negative. - For \( x \in (-8,0) \) (e.g., \( x=-4 \)): - Numerator: \( 3(-4)=-12 \) (negative). - Denominator: \[ (-4-7)(-4+8)=(-11)(4)=-44 \quad \text{(negative)}. \] - Negative divided by negative yields positive, so \( h(x) \) is positive. - For \( x \in (0,7) \) (e.g., \( x=1 \)): - Numerator: \( 3(1)=3 \) (positive). - Denominator: \[ (1-7)(1+8)=(-6)(9)=-54 \quad \text{(negative)}. \] - Positive divided by negative yields negative, so \( h(x) \) is negative. - For \( x \in (7,\infty) \) (e.g., \( x=10 \)): - Numerator: \( 3(10)=30 \) (positive). - Denominator: \[ (10-7)(10+8)=(3)(18)=54 \quad \text{(positive)}. \] - Positive divided by positive is positive, so \( h(x) \) is positive. 5. Analyze the equality condition \( h(x) \geq 0 \): - \( h(x) \) is positive for \( x \in (-8,0) \) and \( x \in (7,\infty) \). - The function equals zero when the numerator is zero (and denominator is nonzero), which occurs at \( x=0 \). Since \( x=0 \) lies in the interval \( (-8,0) \), it can be included. 6. Write the solution set, keeping in mind that \( x=-8 \) and \( x=7 \) are not in the domain: \[ x\in (-8,0] \cup (7,\infty) \] Thus, the solution is: \[ \boxed{(-8,0] \cup (7,\infty)} \]