Question

Use algebraic​ and/or graphical methods to solve the inequality.
negative x cubed minus 6 x squared minus 9 xless than or equals0
Question content area bottom
Part 1
Choose the correct answer below.
A.
The solutions are the values of x in the interval left bracket 0 comma infinity right parenthesis as well as the point xequalsminus3.
B.
The solution is xequalsminus3.
C.
The solutions are the values of x in the interval left bracket 0 comma infinity right parenthesis.
D.
There is no solution.

Answer 1

To solve the inequality \(-x^3 - 6x^2 - 9x \leq 0\), we start by rewriting it:

\[
-x^3 - 6x^2 - 9x \leq 0
\]

We can factor out a negative sign from the inequality:

\[
-(x^3 + 6x^2 + 9x) \leq 0
\]

This simplifies to:

\[
x^3 + 6x^2 + 9x \geq 0
\]

Next, we factor the cubic polynomial. First, we can factor out an \(x\):

\[
x(x^2 + 6x + 9) \geq 0
\]

Now, the quadratic \(x^2 + 6x + 9\) can be factored further:

\[
x^2 + 6x + 9 = (x + 3)^2
\]

So, we can rewrite our inequality as:

\[
x(x + 3)^2 \geq 0
\]

Next, we find the critical points by setting each factor equal to zero:

1. \(x = 0\)
2. \((x + 3)^2 = 0 \Rightarrow x = -3\)

The critical points are \(x = 0\) and \(x = -3\). Now, we will test the intervals defined by these critical points: \( (-\infty, -3) \), \( (-3, 0) \), and \( (0, \infty) \).

1. **Interval** \( (-\infty, -3) \):
- Choose \(x = -4\):
\[
(-4)((-4) + 3)^2 = (-4)(-1)^2 = -4 < 0 \quad \text{(not a solution)}
\]

2. **Interval** \( (-3, 0) \):
- Choose \(x = -2\):
\[
(-2)((-2) + 3)^2 = (-2)(1)^2 = -2 < 0 \quad \text{(not a solution)}
\]

3. **Interval** \( (0, \infty) \):
- Choose \(x = 1\):
\[
(1)((1) + 3)^2 = (1)(4)^2 = 16 > 0 \quad \text{(solution)}
\]

We also need to check the endpoints:

- At \(x = -3\):
\[
(-3)((-3) + 3)^2 = (-3)(0)^2 = 0 \quad \text{(included)}
\]

- At \(x = 0\):
\[
(0)((0) + 3)^2 = (0)(3)^2 = 0 \quad \text{(included)}
\]

From this analysis, the solution set is:

- \(x \in [0, \infty)\)
- \(x = -3\)

In interval notation, the solution is:

\[
[-3, 0) \cup [0, \infty)
\]

So the best answer matching the options provided is **A**. The solutions are the values of \(x\) in the interval \([0, \infty)\) as well as the point \(x = -3\).