How is the range of $x$ determined?

Question

My probability textbook assumes I already understand this but I haven't touched algebra/calc in 5 years before returning to school so I'm confused.

The solution says says for $$0\le y\le(1-x^2).$$

The range of x is: $$-(1-y)^{0.5}\le x\le(1-y)^{0.5}.$$

What exactly should I read up on to understand how this algebraic manipulation works?

Thanks!

Answer 1

we know,

$$x^2=|x|^2$$

but,

$$x^2\leq (1-y)$$

$$|x|^2\leq(1-y)$$

$$|x|\leq(1-y)^{0.5}$$ $$-(1-y)^{0.5}\leq x \leq (1-y)^{0.5}$$

i used this formula if $$|x|\leq k\implies -k\leq x \leq k$$

Answer 2

You need to solve this double inequality for x: $$0\le y\le 1-x^2$$ Let's do $y\le 1-x^2$ first. Adding $x^2$ and subtracting $y$ from both parts, we have $x^2 \le 1-y$. Taking square root from both parts, we obtain: $|x|\le \sqrt{1-y}\:$ or $\:-\sqrt{1-y}\le x \le \sqrt{1-y}\:$ (you know that $\sqrt{1-y}=(1-y)^{0.5}$, right?). We also have that $0\le y \le 1$.

Answer 3

$y\leq(1-x^2)$

$y-1\leq-x^2$

$x^2 \leq 1-y$ (multiplying by a negative number switches the inequality)

$|x| \leq \sqrt{1-y}$ (note that this requires that $y\leq1$)

Recall that |x| means "distance from zero to x". So x is within $ \sqrt{1-y}$ of zero. We can go in either the negative or positive direction, so x can be as small as -$\sqrt{1-y}$ or as large as $\sqrt{1-y}$ i.e.:

-$\sqrt{1-y}\leq x\leq\sqrt{1-y}$