02: Inequalities

1 Following the example in the section $|x| < 3$ leads two cases: $x < 3$ and $-x < 3$. Multiplying both sides of the second case with $-1$ (Rule 3) means that $x > -3$. So the interval of numbers satisfying the given equation is

$$ \phantom{.} \ |x| < 3 \ . $$

2 The inequality leads to two cases: $2x + 1 \leq 1$ and $-(2x + 1) = -2x - 1 \leq 1$. Solving the first case means that

$$ \begin{array}{r c l | r} 2x + 1 & \leq & 1 & -1 \\ 2x & \leq & 1 - 1 = 0 & \div 2 \\ x & \leq & 0 \end{array} $$

and

$$ \begin{array}{r c l | r} -2x - 1 & \leq & 1 & +1 \\ -2x & \leq & 1 + 1 = 2 & \div (-2) \\ x & \leq & -1 \end{array} $$

Gathering the results it can be seen that the numbers satisfying the equation are

$$ \phantom{.} \ x \leq 0 \ . $$

3 As above, the two cases are $x^2 - 2 \leq 1$ and $-(x^2 - 2) = -x^2 + 2 \leq 1$. Solving for $x$ in both cases means adding $2$ to the former case and subtracting $-2$ in the latter. Then both sides of the second case are multiplied by $-1$ and both sides of both cases are squred. This leads to the fact that $x \leq \sqrt{3}$ and $x \geq \sqrt{1}$, so the numbers $x$ satisfyin the inequality are $\sqrt{1} \leq x \leq \sqrt{3}$.

4 Since $|x - 5| = \pm\sqrt{x - 5}^2 = \pm(x-5)$. The inequality $|x - 5| = \pm(x - 5) > 2$ leads to two inequalities, $x - 5 > 2$ and $-(x - 5) = -x + 5 > 2$. Solving for $x$ in both inequalities, means adding $5$ to both sides of the first inequality, subtracting $-5$ from the second inequality and multiplying it by $-1$. Then the solution for the inequality are the number $x > 7$ and numbers $x < -3$.

5 Even though it can be seen from the inequality, letting $y = (x - 1)(x + 2) = x^2 + x - 2$ be equal to $0$, and solving the roots of this quadratic by applying the quadratic formula, the roots of the equation are $x = 1$ or $x = -2$. This divides the $x$-axis into three sections $x < -2$, $-2 < x < 1$, and $x > 1$.

Let $\epsilon > 0$ be arbitrarily close to $0$, then $x - \epsilon \leq x < -2$. Then substituting $x - \epsilon$ in to the inequality leads to inequality $(x - \epsilon)^2 + (x - \epsilon) - 2 < 2$, from where it can be seen that the $(x - \epsilon) - 2$ is very close to zero too, and $(x - \epsilon)^2$ is always nearly $2$ for arbitrarily small $\epsilon$, so $(x - \epsilon)^2$ is always a bit more than $4$, and since $4 > 2$, $x \not < -2$ (that is, $x$ can not be less than $-2$).

Let $x = -\frac{1}{2}$, i.e. the midway at interval $-2 < x < 1$, then

$$ \phantom{,} \ \left( -\frac{1}{2} \right)^2 + \frac{1}{2} - 2 = \dfrac{1}{4} + \dfrac{1}{2} - 2 = \dfrac{5}{4} \ , $$

which is at the interval.

Lastly let $x = 1 + \epsilon$, then substituting $x$ to $x^2 + x - 2$ leads to

$$ \phantom{.} \ (1 + \epsilon)^2 + (1 + \epsilon) - 2 = 1 + 2\epsilon + \epsilon^2 + 1 + \epsilon - 2 = \epsilon^2 + 3 \epsilon \ , $$ $$ \begin{array}{c l} & (1 + \epsilon)^2 + (1 + \epsilon) - 2 \\= & 1 + 2\epsilon + \epsilon^2 + 1 + \epsilon - 2 \\= & \epsilon^2 + 3 \epsilon \ , \end{array} $$

which obviously is not greater than $1$ so $x \not > 1$.

Because $x$ is not less than $-2$, not greater than $1$, and $-2 < x < 1$, the interval for where $x$ is defined, when $(x - 1)(x + 2) < 0$ is the interval

$$ \phantom{.} \ -2 < x < 1 \ . $$