3х²<=18х

To solve the inequality \( 3x^2 \leq 18x \), we can first rearrange it into a standard form: 1. Subtract \( 18x \) from both sides: \[ 3x^2 - 18x \leq 0 \] 2. Factor the left-hand side: \[ 3x(x - 6) \leq 0 \] Now, we will find the critical points by setting the factors equal to zero: 1. \( 3x = 0 \) gives \( x = 0 \) 2. \( x - 6 = 0 \) gives \( x = 6 \) This means the critical points are \( x = 0 \) and \( x = 6 \). Next, we test the intervals defined by these critical points: 1. Interval \( (-\infty, 0) \) 2. Interval \( (0, 6) \) 3. Interval \( (6, +\infty) \) **Testing intervals:** - For the interval \( (-\infty, 0) \), choose \( x = -1 \): \[ 3(-1)(-1 - 6) = 3(-1)(-7) = 21 \quad \text{(positive)} \] - For the interval \( (0, 6) \), choose \( x = 1 \): \[ 3(1)(1 - 6) = 3(1)(-5) = -15 \quad \text{(negative)} \] - For the interval \( (6, +\infty) \), choose \( x = 7 \): \[ 3(7)(7 - 6) = 3(7)(1) = 21 \quad \text{(positive)} \] **Summary of signs:** - \( (-\infty, 0) \): positive - \( (0, 6) \): negative - \( (6, +\infty) \): positive The inequality \( 3x(x - 6) \leq 0 \) is satisfied in the interval where the expression is negative, which is \( (0, 6) \), and also at the points where the expression is zero, which are \( x = 0 \) and \( x = 6 \). **Final solution:** \[ x \in [0, 6] \]