Finite Math Examples

$y = x^{2} - 2 x - 8$

Step 1

Interchange the variables.

$x = y^{2} - 2 y - 8$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $y^{2} - 2 y - 8 = x$ .

$y^{2} - 2 y - 8 = x$

Step 2.2

Subtract $x$ from both sides of the equation.

$y^{2} - 2 y - 8 - x = 0$

Step 2.3

Use the quadratic formula to find the solutions.

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

Step 2.4

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 8 - x$ into the quadratic formula and solve for $y$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot (- 8 - x))}}{2 \cdot 1}$

Step 2.5

Simplify.

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 8 - x)}}{2 \cdot 1}$

Step 2.5.1.2

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (- 8 - x)}}{2 \cdot 1}$

Step 2.5.1.3

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 8 - 4 (- x)}}{2 \cdot 1}$

Step 2.5.1.4

Multiply $- 4$ by $- 8$ .

$y = \frac{2 \pm \sqrt{4 + 32 - 4 (- x)}}{2 \cdot 1}$

Step 2.5.1.5

Multiply $- 1$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 + 32 + 4 x}}{2 \cdot 1}$

Step 2.5.1.6

Add $4$ and $32$ .

$y = \frac{2 \pm \sqrt{36 + 4 x}}{2 \cdot 1}$

Step 2.5.1.7

Factor $4$ out of $36 + 4 x$ .

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Step 2.5.1.7.1

Factor $4$ out of $36$ .

$y = \frac{2 \pm \sqrt{4 \cdot 9 + 4 x}}{2 \cdot 1}$

Step 2.5.1.7.2

Factor $4$ out of $4 \cdot 9 + 4 x$ .

$y = \frac{2 \pm \sqrt{4 (9 + x)}}{2 \cdot 1}$

Step 2.5.1.8

Rewrite $4 (9 + x)$ as $2^{2} (3^{2} + x)$ .

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Step 2.5.1.8.1

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

$y = \frac{2 \pm \sqrt{2^{2} (9 + x)}}{2 \cdot 1}$

Step 2.5.1.8.2

Rewrite $9$ as $3^{2}$ .

$y = \frac{2 \pm \sqrt{2^{2} (3^{2} + x)}}{2 \cdot 1}$

Step 2.5.1.9

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{3^{2} + x}}{2 \cdot 1}$

Step 2.5.1.10

Raise $3$ to the power of $2$ .

$y = \frac{2 \pm 2 \sqrt{9 + x}}{2 \cdot 1}$

Step 2.5.2

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{9 + x}}{2}$

Step 2.5.3

Simplify $\frac{2 \pm 2 \sqrt{9 + x}}{2}$ .

$y = 1 \pm \sqrt{9 + x}$

Step 2.6

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

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Step 2.6.1

Simplify the numerator.

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Step 2.6.1.1

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 8 - x)}}{2 \cdot 1}$

Step 2.6.1.2

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (- 8 - x)}}{2 \cdot 1}$

Step 2.6.1.3

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 8 - 4 (- x)}}{2 \cdot 1}$

Step 2.6.1.4

Multiply $- 4$ by $- 8$ .

$y = \frac{2 \pm \sqrt{4 + 32 - 4 (- x)}}{2 \cdot 1}$

Step 2.6.1.5

Multiply $- 1$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 + 32 + 4 x}}{2 \cdot 1}$

Step 2.6.1.6

Add $4$ and $32$ .

$y = \frac{2 \pm \sqrt{36 + 4 x}}{2 \cdot 1}$

Step 2.6.1.7

Factor $4$ out of $36 + 4 x$ .

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Step 2.6.1.7.1

Factor $4$ out of $36$ .

$y = \frac{2 \pm \sqrt{4 \cdot 9 + 4 x}}{2 \cdot 1}$

Step 2.6.1.7.2

Factor $4$ out of $4 \cdot 9 + 4 x$ .

$y = \frac{2 \pm \sqrt{4 (9 + x)}}{2 \cdot 1}$

Step 2.6.1.8

Rewrite $4 (9 + x)$ as $2^{2} (3^{2} + x)$ .

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Step 2.6.1.8.1

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

$y = \frac{2 \pm \sqrt{2^{2} (9 + x)}}{2 \cdot 1}$

Step 2.6.1.8.2

Rewrite $9$ as $3^{2}$ .

$y = \frac{2 \pm \sqrt{2^{2} (3^{2} + x)}}{2 \cdot 1}$

Step 2.6.1.9

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{3^{2} + x}}{2 \cdot 1}$

Step 2.6.1.10

Raise $3$ to the power of $2$ .

$y = \frac{2 \pm 2 \sqrt{9 + x}}{2 \cdot 1}$

Step 2.6.2

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{9 + x}}{2}$

Step 2.6.3

Simplify $\frac{2 \pm 2 \sqrt{9 + x}}{2}$ .

$y = 1 \pm \sqrt{9 + x}$

Step 2.6.4

Change the $\pm$ to $+$ .

$y = 1 + \sqrt{9 + x}$

Step 2.7

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

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Step 2.7.1

Simplify the numerator.

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Step 2.7.1.1

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

$y = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot (- 8 - x)}}{2 \cdot 1}$

Step 2.7.1.2

Multiply $- 4$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 4 \cdot (- 8 - x)}}{2 \cdot 1}$

Step 2.7.1.3

Apply the distributive property.

$y = \frac{2 \pm \sqrt{4 - 4 \cdot - 8 - 4 (- x)}}{2 \cdot 1}$

Step 2.7.1.4

Multiply $- 4$ by $- 8$ .

$y = \frac{2 \pm \sqrt{4 + 32 - 4 (- x)}}{2 \cdot 1}$

Step 2.7.1.5

Multiply $- 1$ by $- 4$ .

$y = \frac{2 \pm \sqrt{4 + 32 + 4 x}}{2 \cdot 1}$

Step 2.7.1.6

Add $4$ and $32$ .

$y = \frac{2 \pm \sqrt{36 + 4 x}}{2 \cdot 1}$

Step 2.7.1.7

Factor $4$ out of $36 + 4 x$ .

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Step 2.7.1.7.1

Factor $4$ out of $36$ .

$y = \frac{2 \pm \sqrt{4 \cdot 9 + 4 x}}{2 \cdot 1}$

Step 2.7.1.7.2

Factor $4$ out of $4 \cdot 9 + 4 x$ .

$y = \frac{2 \pm \sqrt{4 (9 + x)}}{2 \cdot 1}$

Step 2.7.1.8

Rewrite $4 (9 + x)$ as $2^{2} (3^{2} + x)$ .

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Step 2.7.1.8.1

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

$y = \frac{2 \pm \sqrt{2^{2} (9 + x)}}{2 \cdot 1}$

Step 2.7.1.8.2

Rewrite $9$ as $3^{2}$ .

$y = \frac{2 \pm \sqrt{2^{2} (3^{2} + x)}}{2 \cdot 1}$

Step 2.7.1.9

Pull terms out from under the radical.

$y = \frac{2 \pm 2 \sqrt{3^{2} + x}}{2 \cdot 1}$

Step 2.7.1.10

Raise $3$ to the power of $2$ .

$y = \frac{2 \pm 2 \sqrt{9 + x}}{2 \cdot 1}$

Step 2.7.2

Multiply $2$ by $1$ .

$y = \frac{2 \pm 2 \sqrt{9 + x}}{2}$

Step 2.7.3

Simplify $\frac{2 \pm 2 \sqrt{9 + x}}{2}$ .

$y = 1 \pm \sqrt{9 + x}$

Step 2.7.4

Change the $\pm$ to $-$ .

$y = 1 - \sqrt{9 + x}$

Step 2.8

The final answer is the combination of both solutions.

$y = 1 + \sqrt{9 + x}$

$y = 1 - \sqrt{9 + x}$

$y = 1 + \sqrt{9 + x}$

$y = 1 - \sqrt{9 + x}$

Step 3

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

$f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$

Step 4

Verify if $f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$ is the inverse of $f (x) = x^{2} - 2 x - 8$ .

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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 $f (x) = x^{2} - 2 x - 8$ and $f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$ and compare them.

Step 4.2

Find the range of $f (x) = x^{2} - 2 x - 8$ .

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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:

$[- 9, \infty)$

Step 4.3

Find the domain of $1 + \sqrt{9 + x}$ .

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Step 4.3.1

Set the radicand in $\sqrt{9 + x}$ greater than or equal to $0$ to find where the expression is defined.

$9 + x \geq 0$

Step 4.3.2

Subtract $9$ from both sides of the inequality.

$x \geq - 9$

Step 4.3.3

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

$[- 9, \infty)$

Step 4.4

Find the domain of $f (x) = x^{2} - 2 x - 8$ .

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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 $f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$ is the range of $f (x) = x^{2} - 2 x - 8$ and the range of $f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$ is the domain of $f (x) = x^{2} - 2 x - 8$ , then $f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$ is the inverse of $f (x) = x^{2} - 2 x - 8$ .

$f^{- 1} (x) = 1 + \sqrt{9 + x}, 1 - \sqrt{9 + x}$

Step 5