Question

Solve the following:

1) x^3 <= 16x
2) (5x/x^2+12x-8) >= 3

Answer 1

1.x^3 - 16x
= x(x^2 - 16)
= x(x^2 + 4x - 4x - 16)
= x[(x^2 + 4x) - (4x + 16)]
= x[x(x + 4) - 4(x + 4)]
= x(x + 4)(x - 4)

Answer 2

1. x^3 - 16x ≤ 0
x(x^2 - 16) ≤ 0
x(x-4)(x+4) ≤ 0

critical values are -4, 0, 4

making a quick sketch of y = x^3 - 16x
shows a standard cubic with x-intercepts of 0, ± 4
Where is the curve below the x-axis ?

x ≤ -4 or 0 ≤ x ≤ 4

2.
(5x/x^2+12x-8) >= 3
5x > 3x^2 + 36x - 24
3x^2 + 31x - 24 < 0

consider 3x^2 + 31x - 24
x = (-31 ± √1249)/6
= appr .7235 or -11.057

critical values: x = .7235 , x = -11.057

We also have to consider
x^2 + 12x - 8 , it cannot be zero
let's see for what values it is
solving, I got x = .63325 , x = -12.6332

looking at Wolfram's graph:
http://www.wolframalpha.com/input/?i=5x%2F(x%5E2%2B12x-8)+%3E%3D+3

we have
-12.6332 < x ≤ -11.057
OR
.63325 < x ≤ .7235