Trigonometry Examples

$f (x) = \frac{{(x + 7)}^{2}}{6}$

Step 1

Write $f (x) = \frac{{(x + 7)}^{2}}{6}$ as an equation.

$y = \frac{{(x + 7)}^{2}}{6}$

Step 2

Interchange the variables.

$x = \frac{{(y + 7)}^{2}}{6}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{{(y + 7)}^{2}}{6} = x$ .

$\frac{{(y + 7)}^{2}}{6} = x$

Step 3.2

Multiply both sides of the equation by $6$ .

$6 \frac{{(y + 7)}^{2}}{6} = 6 x$

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $6$ .

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Step 3.3.1.1

Cancel the common factor.

$6 \frac{{(y + 7)}^{2}}{6} = 6 x$

Step 3.3.1.2

Rewrite the expression.

${(y + 7)}^{2} = 6 x$

Step 3.4

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

$y + 7 = \pm \sqrt{6 x}$

Step 3.5

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

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Step 3.5.1

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

$y + 7 = \sqrt{6 x}$

Step 3.5.2

Subtract $7$ from both sides of the equation.

$y = \sqrt{6 x} - 7$

Step 3.5.3

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

$y + 7 = - \sqrt{6 x}$

Step 3.5.4

Subtract $7$ from both sides of the equation.

$y = - \sqrt{6 x} - 7$

Step 3.5.5

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

$y = \sqrt{6 x} - 7$

$y = - \sqrt{6 x} - 7$

$y = \sqrt{6 x} - 7$

$y = - \sqrt{6 x} - 7$

$y = \sqrt{6 x} - 7$

$y = - \sqrt{6 x} - 7$

Step 4

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

$f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$

Step 5

Verify if $f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$ is the inverse of $f (x) = \frac{{(x + 7)}^{2}}{6}$ .

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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 $f (x) = \frac{{(x + 7)}^{2}}{6}$ and $f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$ and compare them.

Step 5.2

Find the range of $f (x) = \frac{{(x + 7)}^{2}}{6}$ .

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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:

$[0, \infty)$

Step 5.3

Find the domain of $\sqrt{6 x} - 7$ .

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Step 5.3.1

Set the radicand in $\sqrt{6 x}$ greater than or equal to $0$ to find where the expression is defined.

$6 x \geq 0$

Step 5.3.2

Divide each term in $6 x \geq 0$ by $6$ and simplify.

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Step 5.3.2.1

Divide each term in $6 x \geq 0$ by $6$ .

$\frac{6 x}{6} \geq \frac{0}{6}$

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Cancel the common factor of $6$ .

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Step 5.3.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} \geq \frac{0}{6}$

Step 5.3.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{0}{6}$

Step 5.3.2.3

Simplify the right side.

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Step 5.3.2.3.1

Divide $0$ by $6$ .

$x \geq 0$

Step 5.3.3

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

$[0, \infty)$

Step 5.4

Find the domain of $f (x) = \frac{{(x + 7)}^{2}}{6}$ .

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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 $f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$ is the range of $f (x) = \frac{{(x + 7)}^{2}}{6}$ and the range of $f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$ is the domain of $f (x) = \frac{{(x + 7)}^{2}}{6}$ , then $f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$ is the inverse of $f (x) = \frac{{(x + 7)}^{2}}{6}$ .

$f^{- 1} (x) = \sqrt{6 x} - 7, - \sqrt{6 x} - 7$

Step 6