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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?

State the domain: .[tex]y \:=\: x\sqrt{1 - x^2}[/tex]

So: .[tex]1 - x^2 \:\ge\:0[/tex]

. . . . . [tex]-x^2 \:\ge\:-1[/tex]

. . . . . . .[tex]x^2\:\le\:1[/tex]

[tex]\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}[/tex]

- Jan 29, 2012

- 1,151

NOT \(\displaystyle \{-1, 1\}\) because that means the numbers -1 and 1state 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?

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

- Feb 15, 2012

- 1,967

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:

$|x| \leq 1$.