Algebra Examples

$f (x) = {(x - 5)}^{2}$

Step 1

Write $f (x) = {(x - 5)}^{2}$ as an equation.

$y = {(x - 5)}^{2}$

Step 2

Interchange the variables.

$x = {(y - 5)}^{2}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as ${(y - 5)}^{2} = x$ .

${(y - 5)}^{2} = x$

Step 3.2

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

$y - 5 = \pm \sqrt{x}$

Step 3.3

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

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Step 3.3.1

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

$y - 5 = \sqrt{x}$

Step 3.3.2

Add $5$ to both sides of the equation.

$y = \sqrt{x} + 5$

Step 3.3.3

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

$y - 5 = - \sqrt{x}$

Step 3.3.4

Add $5$ to both sides of the equation.

$y = - \sqrt{x} + 5$

Step 3.3.5

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

$y = \sqrt{x} + 5$

$y = - \sqrt{x} + 5$

$y = \sqrt{x} + 5$

$y = - \sqrt{x} + 5$

$y = \sqrt{x} + 5$

$y = - \sqrt{x} + 5$

Step 4

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

$f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$

Step 5

Verify if $f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$ is the inverse of $f (x) = {(x - 5)}^{2}$ .

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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 $f (x) = {(x - 5)}^{2}$ and $f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$ and compare them.

Step 5.2

Find the range of $f (x) = {(x - 5)}^{2}$ .

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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 $\sqrt{x} + 5$ .

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Step 5.3.1

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

$x \geq 0$

Step 5.3.2

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

$[0, \infty)$

Step 5.4

Find the domain of $f (x) = {(x - 5)}^{2}$ .

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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 $f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$ is the range of $f (x) = {(x - 5)}^{2}$ and the range of $f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$ is the domain of $f (x) = {(x - 5)}^{2}$ , then $f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$ is the inverse of $f (x) = {(x - 5)}^{2}$ .

$f^{- 1} (x) = \sqrt{x} + 5, - \sqrt{x} + 5$

Step 6