Problem 4

Find all numbers $x$ for which these assertions are true.

(i) \[ \begin{align*} 4 - x &< 3 - 2x \\ x &< -1 \end{align*} \]

(ii) \[ \begin{align*} 5 - x^2 &< 8 \\ x^2 &> -3 \end{align*} \] We can dinstinguish that in 3 cases. First $x > 0$: \[ \begin{align*} x^2 \in P \rightarrow x^2 > -3 \end{align*} \] Second: $x = 0$: \[ \begin{align*} 0 > -3 \end{align*} \] And finally $x < 0$: \[ \begin{align*} (-x)(-x) &= x^2 \in P \\ &\rightarrow x^2 > -3 \end{align*} \]

(iii) \[ \begin{align*} 5 - x^2 &< -2 \\ -x^2 + 7 &< 0 \\ (x + \sqrt{7})(-x + \sqrt{7}) &< 0 \\ \end{align*} \] That leads us to 2 cases. First: \[ \begin{align*} x + \sqrt{7} &< 0 \\ x &< -\sqrt{7} \end{align*} \] Second: \[ \begin{align*} -x + \sqrt{7} &< 0 \\ x &> \sqrt{7} \end{align*} \]

(iv) \[ \begin{align*} (x - 1)(x - 3) &> 0 \\ \\ (x - 1) &> 0 && \text{| firse case} \\ x &> 1 \\ \\ (x - 3) &> 0 && \text{| second case}\\ x &> 3 \\ \\ \rightarrow x &> 3 > 1 \end{align*} \]

(v) \[ \begin{align*} x^2 - 2x + 2 &> 0 \\ (x - 1)^2 + 1 &> 0 \\ \\ (x - 1)^2 &> -1 && \text{| with } y = x - 1 \\ y^2 &> -1 && \text{| equal to (ii)} \end{align*} \]

(vi) \[ \begin{align*} x^2 + x + 1 &> 2 \\ x^2 + x - 1 &> 0 \\ (x + \frac{1}{2})^2 - (\frac{1}{2})^2 - 1 &> 0 \\ (x + \frac{1}{2})^2 &> \frac{5}{4} \\ \sqrt{(x + \frac{1}{2})^2} &> \sqrt{\frac{5}{4}} \\ \pm(x + \frac{1}{2}) &> \frac{\sqrt{5}}{2} \end{align*} \] First case: \[ \begin{align*} -x - \frac{1}{2} &> \frac{\sqrt{5}}{2} \\ x &< -\frac{1 - \sqrt{5}}{2} \end{align*} \] Second case: \[ \begin{align*} x + \frac{1}{2} &> \frac{\sqrt{5}}{2} \\ x &> -\frac{1 + \sqrt{5}}{2} \end{align*} \]

(vii) \[ \begin{align*} x^2 - x + 10 &> 16 \\ x^2 - x - 6 &> 0 \\ (x - \frac{1}{2})^2 &> \frac{25}{4} \\ \pm (x - \frac{1}{2}) &> \frac{5}{2} \end{align*} \] First case: \[ \begin{align*} x - \frac{1}{2} &> \frac{5}{2} \\ x &> 3 \end{align*} \] Second case: \[ \begin{align*} - x + \frac{1}{2} &> \frac{5}{2} \\ x &< -2 \end{align*} \]

(viii) \[ \begin{align*} x^2 + x + 1 &> 0 \\ (x + \frac{1}{2})^2 - \frac{1}{4} + 1 &> 0 \\ \pm (x + \frac{1}{2}) &> \frac{\sqrt{-3}}{2} && \text{| complex numbers!} \end{align*} \] We can by pass the complex numbers by rephrasing the problem to $x^2 + x > -1$ and check if there is a scenario where the left expression ever becomes negative. For $x \geq 0 \rightarrow x^2 + x > 0$ and for $x < 0 \rightarrow x^2 \in P$. The expression then never reaches $-1$. That means, every $x$ is possible.

(ix) \[ \begin{align*} (x - \pi)(x + 5)(x - 3) &> 0 \end{align*} \] From that:


Hence, $x > \pi$.

(x) \[ \begin{align*} (x - \sqrt[3]{2})(x - \sqrt{2}) &> 0 \\ \\ x &> \sqrt{2} && \text{| conclusion from 1. and 2. case} \end{align*} \]

(xi) \[ \begin{align*} 2^x &< 8 \\ x &< log_{2}8 \\ x &< 3 \end{align*} \]

(xii) \[ \begin{align*} x + 3^x < 4 \end{align*} \] This formular I wouldn't know yet how to solve. Looking at the left side the whole expression becomes equal or larger than 4 when $x >= 1$ so $x < 1$.

(xiii) \[ \begin{align*} \frac{1}{x} + \frac{1}{1 - x} &> 0 \end{align*} \] $x$ isn't allowed to reach 2 value in this inequality: $x = 0$ and $x = 1$. Both lead to a division by 0. When $x < 0$ then $-\frac{1}{x} + \frac{1}{1 + x} < 0$. When $x > 1$ then $\frac{1}{x} - \frac{1}{x - 1} < 0$. But we are left with $0 < x < 1$ where $\frac{1}{x}, \frac{1}{1 - x} \in P$

(xiv) \[ \begin{align*} \frac{x - 1}{x + 1} &> 0 \end{align*} \] We only have to know when the nominator and denominator are in $P$. First case: \[ \begin{align*} x - 1 &> 0 \\ x &> 1 \end{align*} \] Second case: \[ \begin{align*} x + 1 &> 0 \\ x &> -1 \end{align*} \] From that follows $x > 1$.


You found a typo or some other mistake I made in this text? All articles can be changed here. If you want to exchange ideas then simply drop me a message at contact@paulheymann.de.