## Monotonicity with a parameter

### Task number: 2875

Determine for which real numbers \(x\) the following sequence is monotonic.

\(\displaystyle \left\{ \left( \frac{x^3}{3x-2} \right)^n \right\} \)

#### Hint

Consider how the monotonicity of the sequence \(\{a^n\}\) depends on the parameter \(a\).

#### Resolution

The sequence is undefined for \(x=\frac23\).

The sequence is constant if \(\displaystyle \frac{x^3}{3x-2}=0 \) or \(\displaystyle \frac{x^3}{3x-2}=1 \).

In the first case \(x=0\), in the second \(x\) is a root of the cubic equation \(x^3-3x+2=0\), so \(x=1\) (a double root) or \(x=-2\).

The sequence is alternating if \(\displaystyle \frac{x^3}{3x-2}<0 \). If the numerator is negative, then the denominator must be negative as well. So the remaining possibility is that the numerator is positive and the denominator is negative, which directly gives us the interval \(\left(0,\frac23\right)\).

The sequence is decreasing for \(\displaystyle 0<\frac{x^3}{3x-2}<1 \).

For \(x<0\) we solve the inequality \(x^3>3x-2\), which holds in the interval \((-2{,}0)\).

For \(x>0\) we solve the inequality \(x^3<3x-2\), which has no solution.

In every other case, \(\displaystyle \frac{x^3}{3x-2}>1 \).

We can verify this by case analysis, because \(x^3<3x-2\) on the interval \((-\infty,-2)\) , and \(x^3\ge 3x-2\) on the interval \(\left(\frac23,\infty\right)\) . Furthermore, equality holds only for \(x=1\), otherwise the inequality is strict.

#### Result

The sequences is \( \begin{cases} \text{ constant } & \text{ for } x\in \{-2{,}0,1\},\\ \text{ increasing } & \text{ for } x\in (-\infty,-2)\cup \left(\frac23{,}1\right)\cup(1,\infty),\\ \text{ decreasing } & \text{ for } x\in (-2{,}0). \end{cases} \)