How would i find the solutions to x^3 - 9x > 0?

Question

it is actually greater than or equal to zero, but i dont know how to subscript >_.

Answer 1

Greetings,

x^3 - 9x ≥ 0

x(x^2 - 9) ≥ 0

x(x - 3)(x + 3) ≥ 0

------o+++++ x

----------o+++ (x - 3)

---o++++++ (x + 3)

..-3..0..3

The overall product is positive when all 3 factors are positive or when 2 factors are negative and the third positive

ie x ≥ 3 or -3 ≤ x ≤ 0

Regards

Answer 2

x³ - 9x ≥ 0   ...factor out an x

x(x² - 9) ≥ 0   ...factor x² - 9

x(x - 3)(x + 3) ≥ 0

The zeros of x(x - 3)(x + 3) are at x = 0, x = -3 and x = 3.  Since the function is x³ - 9x, it heads to negative infinity as x goes to negative infinity and heads to positive infinity as x goes to positive infinity.  Then the function is greater than zero for x inbetween -3 and 0 and greater than 3.  

Solutions:  -3 ≤ x ≤ 0, or x ≥ 3

Hope this helps you!