Find all solutions to the inequality \[\sqrt[3]{x}+ \frac{2}{\sqrt[3]{x}+ 3} \le 0,\]giving your answer in interval notation.
Seeing the expression $\sqrt[3]{x}$ twice, we make the substitution $y = \sqrt[3]{x},$ so that our inequality becomes \[y + \frac{2}{y+3} \le 0.\]Combining the terms on the left-hand side under a common denominator, we get \[\frac{y^2+3y+2}{y+3} \le 0,\]which factors as \[\frac{(y+1)(y+2)}{y+3} \le 0.\]Letting $f(y) = (y+1)(y+2)/(y+3),$ we make a sign table based on this inequality: \begin{tabular}{c|ccc|c} &$y+1$ &$y+2$ &$y+3$ &$f(y)$ \\ \hline$y<-3$ &$-$&$-$&$-$&$-$\\ [.1cm]$-3<y<-2$ &$-$&$-$&$+$&$+$\\ [.1cm]$-2<y<-1$ &$-$&$+$&$+$&$-$\\ [.1cm]$y>-1$ &$+$&$+$&$+$&$+$\\ [.1cm]\end{tabular}Therefore, the inequality holds if $y < -3$ or $-2 < y < -1.$ Since the inequality is nonstrict, we must also include the values of $y$ that make $f(y) = 0,$ which are $y=-1$ and $y=-2.$ Therefore, the solutions to this inequality are \[y \in (-\infty, -3) \cup [-2, -1].\]Since $y = \sqrt[3]{x},$ we have either $\sqrt[3]{x} < -3$ or $-2 \le \sqrt[3]{x} \le -1.$ Since $\sqrt[3]{x}$ is an increasing function of $x,$ we can cube all sides of these inequalities, to get $x < -27$ and $-8 \le x \le -1,$ respectively. Therefore, \[x \in \boxed{(-\infty, -27) \cup [-8, -1]}.\]
(-\iny,-27)\cup[-8,-1]
Level 5
Intermediate Algebra
Problem: Find all solutions to the inequality \[\sqrt[3]{x}+ \frac{2}{\sqrt[3]{x}+ 3} \le 0,\]giving your answer in interval notation.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality \[\sqrt[3]{x}+ \frac{2}{\sqrt[3]{x}+ 3} \le 0,\] we first define \(y = \sqrt[3]{x}\). This transforms the inequality into \(y + \frac{2}{y + 3} \le 0\). 

Next, we analyze the expression \(y + \frac{2}{y + 3}\). To combine the terms, we find a common denominator, which is \(y + 3\). Thus, we rewrite the inequality as:

\[
\frac{(y + 3)y + 2}{y + 3} \le 0.
\]

Expanding the numerator gives:

\[
\frac{y^2 + 3y + 2}{y + 3} \le 0.
\]

We can factor the quadratic \(y^2 + 3y + 2\) as \((y + 1)(y + 2)\). Therefore, the inequality becomes:

\[
\frac{(y + 1)(y + 2)}{y + 3} \le 0.
\]

Next, we identify the critical points by setting the numerator and denominator to zero:

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

These critical points divide the number line into intervals: \((-\infty, -3)\), \((-3, -2)\), \((-2, -1)\), and \((-1, \infty)\). We will test each interval to determine where the inequality holds.

1. For \(y < -3\) (e.g., \(y = -4\)):
   \[
   \frac{(-4 + 1)(-4 + 2)}{-4 + 3} = \frac{(-3)(-2)}{-1} = \frac{6}{-1} < 0.
   \]
   This interval satisfies the inequality.

2. For \(-3 < y < -2\) (e.g., \(y = -2.5\)):
   \[
   \frac{(-2.5 + 1)(-2.5 + 2)}{-2.5 + 3} = \frac{(-1.5)(-0.5)}{0.5} = \frac{0.75}{0.5} > 0.
   \]
   This interval does not satisfy the inequality.

3. For \(-2 < y < -1\) (e.g., \(y = -1.5\)):
   \[
   \frac{(-1.5 + 1)(-1.5 + 2)}{-1.5 + 3} = \frac{(-0.5)(0.5)}{1.5} = \frac{-0.25}{1.5} < 0.
   \]
   This interval satisfies the inequality.

4. For \(y > -1\) (e.g., \(y = 0\)):
   \[
   \frac{(0 + 1)(0 + 2)}{0 + 3} = \frac{(1)(2)}{3} > 0.
   \]
   This interval does not satisfy the inequality.

Next, we check the critical points:

- At \(y = -3\), the expression is undefined.
- At \(y = -2\), the expression equals zero.
- At \(y = -1\), the expression equals zero.

Thus, the solution intervals are:

1. From \((-\infty, -3)\) which is included.
2. From \((-2, -1)\) which includes both endpoints.

Now, we convert back to \(x\) using \(y = \sqrt[3]{x}\):

- For \(y < -3\), we have \(x < -27\).
- For \(-2 \leq y \leq -1\), we have \(-8 \leq x \leq -1\).

Combining these results, the final answer in interval notation is:

\[
(-\infty, -27) \cup [-8, -1].
\]<|END_OF_LATENT|> The final answer is $(-\iny,-27)\cup[-8,-1]$.