Algebra Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Subtract $x$ from both sides of the equation.

$y^{2} - 2 y - x = 0$

Step 3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.4

Substitute the values $a = 1$ , $b = - 2$ , and $c = - x$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot (- x))}}{2 \cdot 1}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $- 2$ to the power of $2$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 1 x)}}{2 \cdot 1}$

Step 3.5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 3.5.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (- 1 x)}}{2 \cdot 1}$

Step 3.5.1.2.2

Multiply $- 4$ by $- 1$ .

$y = \frac{2 \pm \sqrt{4 + 4 x}}{2 \cdot 1}$

Step 3.5.1.3

Factor $4$ out of $4 + 4 x$ .

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Step 3.5.1.3.1

Factor $4$ out of $4$ .

$y = \frac{2 \pm \sqrt{4 \cdot 1 + 4 x}}{2 \cdot 1}$

Step 3.5.1.3.2

Factor $4$ out of $4 \cdot 1 + 4 x$ .

$y = \frac{2 \pm \sqrt{4 (1 + x)}}{2 \cdot 1}$

Step 3.5.1.4

Rewrite $4 (1 + x)$ as $2^{2} (1^{2} + x)$ .

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Step 3.5.1.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 3.5.1.4.2

Rewrite $1$ as $1^{2}$ .

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

Step 3.5.1.5

Pull terms out from under the radical.

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

Step 3.5.1.6

One to any power is one.

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

Step 3.5.2

Multiply $2$ by $1$ .

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

Step 3.5.3

Simplify $\frac{2 \pm 2 \sqrt{1 + x}}{2}$ .

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

Step 3.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Raise $- 2$ to the power of $2$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 1 x)}}{2 \cdot 1}$

Step 3.6.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 3.6.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (- 1 x)}}{2 \cdot 1}$

Step 3.6.1.2.2

Multiply $- 4$ by $- 1$ .

$y = \frac{2 \pm \sqrt{4 + 4 x}}{2 \cdot 1}$

Step 3.6.1.3

Factor $4$ out of $4 + 4 x$ .

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Step 3.6.1.3.1

Factor $4$ out of $4$ .

$y = \frac{2 \pm \sqrt{4 \cdot 1 + 4 x}}{2 \cdot 1}$

Step 3.6.1.3.2

Factor $4$ out of $4 \cdot 1 + 4 x$ .

$y = \frac{2 \pm \sqrt{4 (1 + x)}}{2 \cdot 1}$

Step 3.6.1.4

Rewrite $4 (1 + x)$ as $2^{2} (1^{2} + x)$ .

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Step 3.6.1.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 3.6.1.4.2

Rewrite $1$ as $1^{2}$ .

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

Step 3.6.1.5

Pull terms out from under the radical.

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

Step 3.6.1.6

One to any power is one.

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

Step 3.6.2

Multiply $2$ by $1$ .

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

Step 3.6.3

Simplify $\frac{2 \pm 2 \sqrt{1 + x}}{2}$ .

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

Step 3.6.4

Change the $\pm$ to $+$ .

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

Step 3.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

Raise $- 2$ to the power of $2$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 1 x)}}{2 \cdot 1}$

Step 3.7.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 3.7.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (- 1 x)}}{2 \cdot 1}$

Step 3.7.1.2.2

Multiply $- 4$ by $- 1$ .

$y = \frac{2 \pm \sqrt{4 + 4 x}}{2 \cdot 1}$

Step 3.7.1.3

Factor $4$ out of $4 + 4 x$ .

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Step 3.7.1.3.1

Factor $4$ out of $4$ .

$y = \frac{2 \pm \sqrt{4 \cdot 1 + 4 x}}{2 \cdot 1}$

Step 3.7.1.3.2

Factor $4$ out of $4 \cdot 1 + 4 x$ .

$y = \frac{2 \pm \sqrt{4 (1 + x)}}{2 \cdot 1}$

Step 3.7.1.4

Rewrite $4 (1 + x)$ as $2^{2} (1^{2} + x)$ .

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Step 3.7.1.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 3.7.1.4.2

Rewrite $1$ as $1^{2}$ .

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

Step 3.7.1.5

Pull terms out from under the radical.

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

Step 3.7.1.6

One to any power is one.

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

Step 3.7.2

Multiply $2$ by $1$ .

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

Step 3.7.3

Simplify $\frac{2 \pm 2 \sqrt{1 + x}}{2}$ .

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

Step 3.7.4

Change the $\pm$ to $-$ .

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

Step 3.8

The final answer is the combination of both solutions.

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

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

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

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

Step 4

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

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

Step 5

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

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

Step 5.2

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

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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 $1 + \sqrt{1 + x}$ .

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Step 5.3.1

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

$1 + x \geq 0$

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

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

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

Step 6