Solve 3x-x^2leq0

$$3 x - x ^ { 2 } \leq 0$$

Solve for x

$$x\in (-\infty,0]\cup [3,\infty)$$

Solution Steps

Multiply the inequality by -1 to make the coefficient of the highest power in 3x-x^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-3x+x^{2}\geq 0$$

Factor out x.

$$x\left(x-3\right)\geq 0$$

For the product to be ≥0, x and x-3 have to be both ≤0 or both ≥0. Consider the case when x and x-3 are both ≤0.

$$x\leq 0 x-3\leq 0$$

The solution satisfying both inequalities is x\leq 0.

$$x\leq 0$$

Consider the case when x and x-3 are both ≥0.

$$x-3\geq 0 x\geq 0$$

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\leq 0\text{; }x\geq 3$$