I have to solve the inequality:
$$\sqrt{4-x^2} > 1 – x$$
in $\mathbb{R}$. Obviously we have to apply the condition of $x \in (-\infty, 2] \cup [2, +\infty)$ for the square root to be defined. But after that I have a few problems.
There is no problem for the case when we have $1-x \ge 0$ since both sides of the inequality would be positive so I could just square both sides of the inequality (since the inequality sign would not change) and find an answer. However, the case when we have $1-x < 0$ (so $x \in (1, +\infty)$) confuses me. When does the inequality sign change and when does it not? Or if that is not how I should I approach this case, how should I?
Best Answer
For the square root to be defined we need $x^{2} \leq 4$ or $-2 \leq x \leq 2$. If $1-x <0$ then the given inequality holds automatically. So very number in $(1,2]$ is a solution. For $-2 \leq x <1$ we have $4-x^{2} >(1-x)^{2}=x^{2}-2x+1$ or $2x^{2}-2x-3<0$. This can be written as $(x-\frac 1 2 )^{2} <7/4$. Can you finish?