\( \frac { a x } { x + 1 } \leq 0 \)

To solve the inequality \( \frac{ax}{x+1} \leq 0 \), we need to consider the signs of the numerator \( ax \) and the denominator \( x+1 \) separately, and then combine the results. First, let's find the critical points where the signs of the numerator and denominator could change. These are the points where \( ax = 0 \) and \( x+1 = 0 \). 1. For \( ax = 0 \), we have \( x = 0 \) (assuming \( a \neq 0 \)). 2. For \( x+1 = 0 \), we have \( x = -1 \). Now, let's consider the intervals created by these critical points: \( (-\infty, -1) \), \( (-1, 0) \), and \( (0, \infty) \). We will test a value in each interval to determine the sign of \( \frac{ax}{x+1} \) in that interval: 1. For \( x \in (-\infty, -1) \), let's test \( x = -2 \): \[ \frac{a(-2)}{-2+1} = \frac{-2a}{-1} = 2a \] The sign of \( 2a \) depends on the sign of \( a \). If \( a > 0 \), then \( 2a > 0 \); if \( a < 0 \), then \( 2a < 0 \). 2. For \( x \in (-1, 0) \), let's test \( x = -0.5 \): \[ \frac{a(-0.5)}{-0.5+1} = \frac{-0.5a}{0.5} = -a \] The sign of \( -a \) depends on the sign of \( a \). If \( a > 0 \), then \( -a < 0 \); if \( a < 0 \), then \( -a > 0 \). 3. For \( x \in (0, \infty) \), let's test \( x = 1 \): \[ \frac{a(1)}{1+1} = \frac{a}{2} \] The sign of \( \frac{a}{2} \) depends on the sign of \( a \). If \( a > 0 \), then \( \frac{a}{2} > 0 \); if \( a < 0 \), then \( \frac{a}{2} < 0 \). Now, let's consider the inequality \( \frac{ax}{x+1} \leq 0 \): - If \( a > 0 \), then the inequality is true when \( x \) is in the intervals \( (-\infty, -1) \) and \( (0, \infty) \), because in both intervals, the numerator \( ax \) is positive and the denominator \( x+1 \) is negative, resulting in a negative value for the fraction. - If \( a < 0 \), then the inequality is true when \( x \) is in the intervals \( (-1, 0) \) and \( (0, \infty) \), because in both intervals, the numerator \( ax \) is negative and the denominator \( x+1 \) is positive, resulting in a negative value for the fraction. - If \( a = 0 \), then the inequality becomes \( 0 \leq 0 \), which is true for all \( x \neq -1 \) (since \( x = -1 \) would make the denominator zero, which is undefined). Therefore, the solution to the inequality \( \frac{ax}{x+1} \leq 0 \) is: - If \( a > 0 \), then \( x \in (-\infty, -1) \cup (0, \infty) \). - If \( a < 0 \), then \( x \in (-1, 0) \cup (0, \infty) \). - If \( a = 0 \), then \( x \in (-\infty, -1) \cup (-1, \infty) \), which simplifies to \( x \neq -1 \).