Graph f(x)=1/( square root of 1-x^2)

$f (x) = \frac{1}{\sqrt{1 - x^{2}}}$

Find the domain for $y = \frac{1}{\sqrt{1 - x^{2}}}$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

Set the radicand in $\sqrt{(1 + x) (1 - x)}$ greater than or equal to $0$ to find where the expression is defined.

$(1 + x) (1 - x) \geq 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$1 + x = 0$

$1 - x = 0$

Set $1 + x$ equal to $0$ and solve for $x$ .

Set $1 + x$ equal to $0$ .

$1 + x = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $1 - x$ equal to $0$ and solve for $x$ .

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The final solution is all the values that make $(1 + x) (1 - x) \geq 0$ true.

$x = - 1, 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$(1 - 4) (1 - (- 4)) \geq 0$

The left side $- 15$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$(1 + 0) (1 - (0)) \geq 0$

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$(1 + 4) (1 - (4)) \geq 0$

The left side $- 15$ is less than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$- 1 \leq x \leq 1$

Set the denominator in $\frac{\sqrt{(1 + x) (1 - x)}}{(1 + x) (1 - x)}$ equal to $0$ to find where the expression is undefined.

$(1 + x) (1 - x) = 0$

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$1 + x = 0$

$1 - x = 0$

Set $1 + x$ equal to $0$ and solve for $x$ .

Set $1 + x$ equal to $0$ .

$1 + x = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $1 - x$ equal to $0$ and solve for $x$ .

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The final solution is all the values that make $(1 + x) (1 - x) = 0$ true.

$x = - 1, 1$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- 1, 1)$

Set-Builder Notation:

${x | - 1 < x < 1}$

Interval Notation:

$(- 1, 1)$

Set-Builder Notation:

${x | - 1 < x < 1}$

To find the end points, substitute the bounds of the $x$ values from the domain into $f (x) = \frac{\sqrt{(1 + x) (1 - x)}}{(1 + x) (1 - x)}$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \frac{\sqrt{(1 - 1) (1 - (- 1))}}{(1 - 1) (1 - (- 1))}$

Remove parentheses.

$f (- 1) = \frac{\sqrt{(1 - 1) (1 - (- 1))}}{(1 - 1) (1 - (- 1))}$

Subtract $1$ from $1$ .

$f (- 1) = \frac{\sqrt{(1 - 1) (1 - (- 1))}}{0 (1 - (- 1))}$

Multiply $- 1$ by $- 1$ .

$f (- 1) = \frac{\sqrt{(1 - 1) (1 - (- 1))}}{0 (1 + 1)}$

Add $1$ and $1$ .

$f (- 1) = \frac{\sqrt{(1 - 1) (1 - (- 1))}}{0 \cdot 2}$

Multiply $0$ by $2$ .

$f (- 1) = \frac{\sqrt{(1 - 1) (1 - (- 1))}}{0}$

The expression contains a division by $0$ . The expression is undefined.

$f (- 1) = Undefined$

Replace the variable $x$ with $1$ in the expression.

$f (1) = \frac{\sqrt{(1 + 1) (1 - (1))}}{(1 + 1) (1 - (1))}$

Remove parentheses.

$f (1) = \frac{\sqrt{(1 + 1) (1 - (1))}}{(1 + 1) (1 - (1))}$

Add $1$ and $1$ .

$f (1) = \frac{\sqrt{(1 + 1) (1 - (1))}}{2 (1 - (1))}$

Multiply $- 1$ by $1$ .

$f (1) = \frac{\sqrt{(1 + 1) (1 - (1))}}{2 (1 - 1)}$

Subtract $1$ from $1$ .

$f (1) = \frac{\sqrt{(1 + 1) (1 - (1))}}{2 \cdot 0}$

Multiply $2$ by $0$ .

$f (1) = \frac{\sqrt{(1 + 1) (1 - (1))}}{0}$

The expression contains a division by $0$ . The expression is undefined.

$f (1) = Undefined$

Undefined

The end points are $(- 1, Undefined), (1, Undefined)$ .

$(- 1, Undefined), (1, Undefined)$

The square root can be graphed using the points around the vertex $(- 1, Undefined), (1, Undefined),$

$\begin{array}{c} x & y \\ - 1 & Undefined \\ 1 & Undefined \end{array}$

Graph f(x)=1/( square root of 1-x^2)

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