Algebra Examples

$y = {(2 x - 1)}^{2} - 2$

Step 1

Interchange the variables.

$x = {(2 y - 1)}^{2} - 2$

Step 2

Solve for $y$ .

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Step 2.1

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

${(2 y - 1)}^{2} - 2 = x$

Step 2.2

Add $2$ to both sides of the equation.

${(2 y - 1)}^{2} = x + 2$

Step 2.3

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

$2 y - 1 = \pm \sqrt{x + 2}$

Step 2.4

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

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Step 2.4.1

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

$2 y - 1 = \sqrt{x + 2}$

Step 2.4.2

Add $1$ to both sides of the equation.

$2 y = \sqrt{x + 2} + 1$

Step 2.4.3

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

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Step 2.4.3.1

Divide each term in $2 y = \sqrt{x + 2} + 1$ by $2$ .

$\frac{2 y}{2} = \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.3.2

Simplify the left side.

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Step 2.4.3.2.1

Cancel the common factor of $2$ .

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Step 2.4.3.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.4

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

$2 y - 1 = - \sqrt{x + 2}$

Step 2.4.5

Add $1$ to both sides of the equation.

$2 y = - \sqrt{x + 2} + 1$

Step 2.4.6

Divide each term in $2 y = - \sqrt{x + 2} + 1$ by $2$ and simplify.

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Step 2.4.6.1

Divide each term in $2 y = - \sqrt{x + 2} + 1$ by $2$ .

$\frac{2 y}{2} = \frac{- \sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.6.2

Simplify the left side.

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Step 2.4.6.2.1

Cancel the common factor of $2$ .

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Step 2.4.6.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{- \sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.6.2.1.2

Divide $y$ by $1$ .

$y = \frac{- \sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.6.3

Simplify the right side.

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Step 2.4.6.3.1

Move the negative in front of the fraction.

$y = - \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

Step 2.4.7

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

$y = \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

$y = - \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

$y = \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

$y = - \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

$y = \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

$y = - \frac{\sqrt{x + 2}}{2} + \frac{1}{2}$

Step 3

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

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

Step 4

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

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

Step 4.2

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

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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:

$[- 2, \infty)$

Step 4.3

Find the domain of $\frac{\sqrt{x + 2}}{2} + \frac{1}{2}$ .

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Step 4.3.1

Set the radicand in $\sqrt{x + 2}$ greater than or equal to $0$ to find where the expression is defined.

$x + 2 \geq 0$

Step 4.3.2

Subtract $2$ from both sides of the inequality.

$x \geq - 2$

Step 4.3.3

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

$[- 2, \infty)$

Step 4.4

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

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

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

Step 5