Question

inequalities question..

Answer 1

\[x ^{3}-81x \le0\]
so i simplified that to:
x(x-9)(x+9) < or = to 0
but i don't know how to put that in interval notation.

Answer 2

@dpaInc how did we do this last time?

Answer 3

\[x^3−81x≤0\]
\[x^2\leq 81\]

Answer 4

i understand how
\[x \le \pm9\] - just i don't get how to write it in interval notation.

Answer 5

@UnkleRhaukus you cant just divide by x o.O

\[x^3 - 81x \le 0\]
\[x(x^2 - 81) \le 0\]
\[x(x+ 9)(x-9) \le 0\]

now equate each factor to 0 (by equate i mean use \(\le\) )

Answer 6

where is x in on the real axis