Question

What is the solution set to `x^3 + x^2 - 4x ≥ 4`?

Answer 1

Solutions are `x in [-2, -1]` and `x in [2, oo)`.

Explanation:

I'm going to assume that you mean to say

`x^3 + x^2 - 4x >= 4`

We can solve this by factoring.

`x^3 + x^2 - 4x - 4 >= 0`

I would write as an equation.

`x^3 + x^2 - 4x -4 = 0`

`x^2(x + 1) - 4(x+ 1) = 0`

`(x^2 - 4)(x + 1) = 0`

`(x + 2)(x - 2)(x + 1) = 0`

`x = -2, 2 and -1`

Now select test points between these values of `x` to see whether they satisfy the initial inequality.

Test point 1: `x = 0`

`0^3 + 0^2 - 4(0) - 4>=^? 0`

`-4 >= 0`

This is obviously false, therefore, `(-1, 2)` obviously is not part of the solution set.

Test point 2: `x = 3`

`3^3 + 3^2 - 4(3) - 4 >=^?0`

`27 + 9 - 12 - 4 >=^? 0`

`20 > 0`

So `[2, oo)` is a solution.

Since `(-1, 2)` is not a solution, by the alternating signs of the function, `[-2, -1]`, is a solution. On the other hand `(-oo, -2)` is not.

Our solutions are: `x in [-2, -1]` and `x in [2, oo)`.

Hopefully this helps!

Answer 2

[-2, -1]
[2, + infinity)

Explanation:

Solving by graphing.
First, graph the function `f(x) = x^3 + x^2 - 4x - 4`
The 3 x-intercepts are: - 2, - 1, and 2
Find parts of the graph that stay above the x-axis , meaning f(x) > 0.
The solution set, where `f(x) >= 0`, are:
Closed interval [-2, - 1],
and half closed interval [2, + infinity)
graph{x^3 + x^2 - 4x - 4 [-5, 5, -2.5, 2.5]}