Chapter 1, Problem 13P

To determine

To solve: The inequality $\ln \left(x^2-2x-2\right)\le 0$.

Answer to Problem 13P

Solution:

The solution is $x\in \left[-1,1-\sqrt{3}\right)\cup \left(1+\sqrt{3},3\right]$$x\in \left[-1,1-\sqrt{3}\right)\cup \left(1+\sqrt{3},3\right]$.

Explanation of Solution

Given:

The inequality: $\ln \left(x^2-2x-2\right)\le 0$.

Calculation:

Consider the inequality $\ln \left(x^2-2x-2\right)\le 0$.

The inequality is not possible as $\ln \left(x^2-2x-2\right)$ value is always positive.

That is, $\ln \left(x^2-2x-2\right)>0$.

Obtain the root of $\left(x^2-2x-2\right)$

$\begin{array}{c}x=\frac{2\pm \sqrt{4+8}}{2}\\ =\frac{2\pm \sqrt{4(3)}}{2}\\ =1\pm \sqrt{3}\end{array}$

Thus the intervals are follows,

$x\in \left(-\infty ,1-\sqrt{3}\right)\cup \left(1+\sqrt{3},\infty \right)$ (1)

Then the inequality,

$\begin{array}{l}\ln \left(x^2-2x-2\right)\le 0\\ x^2-2x-2\le 1\\ x^2-2x-3\le 0\\ \left(x-3\right)\left(x+1\right)\le 0\end{array}$

The roots are $x=3$ and $x=-1$

Thus, the interval as follows,

$x\in \left[-1,3\right]$ (2)

Thus, from the equation (1) and equation (2), it can be concluded that,

$x\in \left[-1,1-\sqrt{3}\right)\cup \left(1+\sqrt{3},3\right]$

Therefore, the solution is $x\in \left[-1,1-\sqrt{3}\right)\cup \left(1+\sqrt{3},3\right]$.