Trigonometry Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

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

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

Step 2.2

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

$y + 2 = \pm \sqrt{x}$

Step 2.3

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

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Step 2.3.1

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

$y + 2 = \sqrt{x}$

Step 2.3.2

Subtract $2$ from both sides of the equation.

$y = \sqrt{x} - 2$

Step 2.3.3

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

$y + 2 = - \sqrt{x}$

Step 2.3.4

Subtract $2$ from both sides of the equation.

$y = - \sqrt{x} - 2$

Step 2.3.5

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

$y = \sqrt{x} - 2$

$y = - \sqrt{x} - 2$

$y = \sqrt{x} - 2$

$y = - \sqrt{x} - 2$

$y = \sqrt{x} - 2$

$y = - \sqrt{x} - 2$

Step 3

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

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

Step 4

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

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

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

Interval Notation:

$[0, \infty)$

Step 4.3

Find the domain of $\sqrt{x} - 2$ .

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Step 4.3.1

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

$x \geq 0$

Step 4.3.2

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

$[0, \infty)$

Step 4.4

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

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Step 4.4.1

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 4.5

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

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

Step 5