Trigonometry Examples

$f (x) = 9 x^{2} - 1$

Step 1

Write $f (x) = 9 x^{2} - 1$ as an equation.

$y = 9 x^{2} - 1$

Step 2

Interchange the variables.

$x = 9 y^{2} - 1$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $9 y^{2} - 1 = x$ .

$9 y^{2} - 1 = x$

Step 3.2

Add $1$ to both sides of the equation.

$9 y^{2} = x + 1$

Step 3.3

Divide each term in $9 y^{2} = x + 1$ by $9$ and simplify.

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Step 3.3.1

Divide each term in $9 y^{2} = x + 1$ by $9$ .

$\frac{9 y^{2}}{9} = \frac{x}{9} + \frac{1}{9}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $9$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{9 y^{2}}{9} = \frac{x}{9} + \frac{1}{9}$

Step 3.3.2.1.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{x}{9} + \frac{1}{9}$

Step 3.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{\frac{x}{9} + \frac{1}{9}}$

Step 3.5

Simplify $\pm \sqrt{\frac{x}{9} + \frac{1}{9}}$ .

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Step 3.5.1

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x + 1}{9}}$

Step 3.5.2

Rewrite $\frac{x + 1}{9}$ as ${(\frac{1}{3})}^{2} (x + 1)$ .

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Step 3.5.2.1

Factor the perfect power $1^{2}$ out of $x + 1$ .

$y = \pm \sqrt{\frac{1^{2} (x + 1)}{9}}$

Step 3.5.2.2

Factor the perfect power $3^{2}$ out of $9$ .

$y = \pm \sqrt{\frac{1^{2} (x + 1)}{3^{2} \cdot 1}}$

Step 3.5.2.3

Rearrange the fraction $\frac{1^{2} (x + 1)}{3^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{1}{3})}^{2} (x + 1)}$

Step 3.5.3

Pull terms out from under the radical.

$y = \pm \frac{1}{3} \sqrt{x + 1}$

Step 3.5.4

Combine $\frac{1}{3}$ and $\sqrt{x + 1}$ .

$y = \pm \frac{\sqrt{x + 1}}{3}$

Step 3.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \frac{\sqrt{x + 1}}{3}$

Step 3.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \frac{\sqrt{x + 1}}{3}$

Step 3.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \frac{\sqrt{x + 1}}{3}$

$y = - \frac{\sqrt{x + 1}}{3}$

$y = \frac{\sqrt{x + 1}}{3}$

$y = - \frac{\sqrt{x + 1}}{3}$

$y = \frac{\sqrt{x + 1}}{3}$

$y = - \frac{\sqrt{x + 1}}{3}$

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$

Step 5

Verify if $f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$ is the inverse of $f (x) = 9 x^{2} - 1$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 9 x^{2} - 1$ and $f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$ and compare them.

Step 5.2

Find the range of $f (x) = 9 x^{2} - 1$ .

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Step 5.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[- 1, \infty)$

Step 5.3

Find the domain of $\frac{\sqrt{x + 1}}{3}$ .

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Step 5.3.1

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

$x + 1 \geq 0$

Step 5.3.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 5.3.3

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

$[- 1, \infty)$

Step 5.4

Find the domain of $f (x) = 9 x^{2} - 1$ .

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Step 5.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 5.5

Since the domain of $f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$ is the range of $f (x) = 9 x^{2} - 1$ and the range of $f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$ is the domain of $f (x) = 9 x^{2} - 1$ , then $f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$ is the inverse of $f (x) = 9 x^{2} - 1$ .

$f^{- 1} (x) = \frac{\sqrt{x + 1}}{3}, - \frac{\sqrt{x + 1}}{3}$

Step 6