Domain of y = x\sqrt{1-x^2}: -1 ≤ x ≤ 1

[shamieh]

state the domain

\(\displaystyle y = x\sqrt{1 - x^2}\)

so

\(\displaystyle 1 - x^2 >= 0\)
\(\displaystyle -x^2 >= -1\)
\(\displaystyle x^2 <= 1\)

\(\displaystyle d = {x <= (+)(-) 1}\) or should I say \(\displaystyle d = {-1, 1}\)

which one is correct?

 

[MarkFL]

I have moved this topic here to our Pre-Calculus subforum as questions on function domains does not involve the calculus. I suspect that this question is part of a calculus question though, or may even come from a calculus textbook, but when choosing the subforum in which to post, the nature of the question itself, rather than from where it originates should be considered firstly.

You are correct in stating:

\(\displaystyle 1-x^2\ge0\)

What I would suggest doing next is factoring the left side:

\(\displaystyle (1+x)(1-x)\ge0\)

Now, determine the critical values, plot them on a number line and test the 3 resulting intervals.

A quicker method would be to plot the expression (the original radicand), and observe where it is non-negative.

What do you find?

 

[soroban]

Hello, shamieh!

State the domain: .$$y \:=\: x\sqrt{1 - x^2}$$

So: .$$1 - x^2 \:\ge\:0$$
. . . . . $$-x^2 \:\ge\:-1$$
. . . . . . .$$x^2\:\le\:1$$

$$\begin{array}{cc}\text{I would write:} & |x| \:\le\:1 \\ & \text{or} \\ & \text{-}1 \:\le\:x\:\le\:1 \\ & \text{or} \\ & [\text{-}1,\,1]\end{array}$$

 

[HallsofIvy]

state the domain

\(\displaystyle y = x\sqrt{1 - x^2}\)

so

\(\displaystyle 1 - x^2 >= 0\)
\(\displaystyle -x^2 >= -1\)
\(\displaystyle x^2 <= 1\)

\(\displaystyle d = {x <= (+)(-) 1}\) or should I say \(\displaystyle d = {-1, 1}\)

which one is correct?

NOT \(\displaystyle \{-1, 1\}\) because that means the numbers -1 and 1 only, not the numbers between. You could say \(\displaystyle \{x| -1\le x\le 1\}\) or [-1, 1] as MarkFL said.

("math" uses the "{" and "}" and as separators itself. To get "{" or "}" in the actual message, use "\{" and "\}".)

 

[Deveno]

A naive approach:

We know that there's "something special" about the points $x = -1, x = 1$. So let's do this:

We'll split the real number line into 5 parts:

Part 1: all numbers less than -1
Part 2: -1
Part 3: all numbers between -1 and 1
Part 4: 1
Part 5: all numbers greater than 1.

Now let's pick numbers in each part, to see what happens:

The 5 real numbers I will use are:

Part 1: -4
Part 2: -1 (no choice, here)
Part 3: 1/2
Part 4: 1 (again, no other choice)
Part 5: 6

Now we will look at $f(x_0)$ for $x_0$ being each one of these 5 numbers:

$f(-4) = 4\sqrt{1 - (-4)^2} = 4\sqrt{-15} = \text{bad}$ (undefined)
$f(-1) = (-1)\sqrt{1 - (-1)^2} = (-1)\sqrt{1 - 1} = (-1)\sqrt{0} = (-1)(0) = 0$ (OK!)
$f(\frac{1}{2}) = \frac{1}{2}\sqrt{1 - (\frac{1}{2})^2} = \frac{1}{2}\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{4}$ (OK!)
$f(1) = (1)\sqrt{1 - 1^2} = (1)(0) = 0$ (OK again!)
$f(6) = 6\sqrt{1 - 6^2} = 6\sqrt{-35} = ?$ (not so good).

So it looks as if what we want is parts 2,3, and 4, and NOT parts 1 and 5. This is:

$\{-1\} \cup (-1,1) \cup \{1\} = [-1,1]$

If you prefer, you can write this as:

$\text{dom}(f) = \{x \in \Bbb R: |x| \leq 1\}$

the absolute value of $x$, written $|x|$ is just another way of saying:

$\sqrt{x^2}$, if one understand square roots as always being non-negative. So if:

$x^2 \leq 1$, then

$\sqrt{x^2} \leq \sqrt{1} = 1$

so:

$|x| \leq 1$.