Finite Math Examples

$f (x) = x^{2} - 5 x + 3$

Step 1

Write $f (x) = x^{2} - 5 x + 3$ as an equation.

$y = x^{2} - 5 x + 3$

Step 2

Interchange the variables.

$x = y^{2} - 5 y + 3$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $y^{2} - 5 y + 3 = x$ .

$y^{2} - 5 y + 3 = x$

Step 3.2

Subtract $x$ from both sides of the equation.

$y^{2} - 5 y + 3 - x = 0$

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 $a = 1$ , $b = - 5$ , and $c = 3 - x$ into the quadratic formula and solve for $y$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot (3 - x))}}{2 \cdot 1}$

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 $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot (3 - x)}}{2 \cdot 1}$

Step 3.5.1.2

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot (3 - x)}}{2 \cdot 1}$

Step 3.5.1.3

Apply the distributive property.

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 3 - 4 (- x)}}{2 \cdot 1}$

Step 3.5.1.4

Multiply $- 4$ by $3$ .

$y = \frac{5 \pm \sqrt{25 - 12 - 4 (- x)}}{2 \cdot 1}$

Step 3.5.1.5

Multiply $- 1$ by $- 4$ .

$y = \frac{5 \pm \sqrt{25 - 12 + 4 x}}{2 \cdot 1}$

Step 3.5.1.6

Subtract $12$ from $25$ .

$y = \frac{5 \pm \sqrt{13 + 4 x}}{2 \cdot 1}$

Step 3.5.2

Multiply $2$ by $1$ .

$y = \frac{5 \pm \sqrt{13 + 4 x}}{2}$

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 $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot (3 - x)}}{2 \cdot 1}$

Step 3.6.1.2

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot (3 - x)}}{2 \cdot 1}$

Step 3.6.1.3

Apply the distributive property.

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 3 - 4 (- x)}}{2 \cdot 1}$

Step 3.6.1.4

Multiply $- 4$ by $3$ .

$y = \frac{5 \pm \sqrt{25 - 12 - 4 (- x)}}{2 \cdot 1}$

Step 3.6.1.5

Multiply $- 1$ by $- 4$ .

$y = \frac{5 \pm \sqrt{25 - 12 + 4 x}}{2 \cdot 1}$

Step 3.6.1.6

Subtract $12$ from $25$ .

$y = \frac{5 \pm \sqrt{13 + 4 x}}{2 \cdot 1}$

Step 3.6.2

Multiply $2$ by $1$ .

$y = \frac{5 \pm \sqrt{13 + 4 x}}{2}$

Step 3.6.3

Change the $\pm$ to $+$ .

$y = \frac{5 + \sqrt{13 + 4 x}}{2}$

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 $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot (3 - x)}}{2 \cdot 1}$

Step 3.7.1.2

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot (3 - x)}}{2 \cdot 1}$

Step 3.7.1.3

Apply the distributive property.

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 3 - 4 (- x)}}{2 \cdot 1}$

Step 3.7.1.4

Multiply $- 4$ by $3$ .

$y = \frac{5 \pm \sqrt{25 - 12 - 4 (- x)}}{2 \cdot 1}$

Step 3.7.1.5

Multiply $- 1$ by $- 4$ .

$y = \frac{5 \pm \sqrt{25 - 12 + 4 x}}{2 \cdot 1}$

Step 3.7.1.6

Subtract $12$ from $25$ .

$y = \frac{5 \pm \sqrt{13 + 4 x}}{2 \cdot 1}$

Step 3.7.2

Multiply $2$ by $1$ .

$y = \frac{5 \pm \sqrt{13 + 4 x}}{2}$

Step 3.7.3

Change the $\pm$ to $-$ .

$y = \frac{5 - \sqrt{13 + 4 x}}{2}$

Step 3.8

The final answer is the combination of both solutions.

$y = \frac{5 + \sqrt{13 + 4 x}}{2}$

$y = \frac{5 - \sqrt{13 + 4 x}}{2}$

$y = \frac{5 + \sqrt{13 + 4 x}}{2}$

$y = \frac{5 - \sqrt{13 + 4 x}}{2}$

Step 4

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

$f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$

Step 5

Verify if $f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$ is the inverse of $f (x) = x^{2} - 5 x + 3$ .

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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 $f (x) = x^{2} - 5 x + 3$ and $f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$ and compare them.

Step 5.2

Find the range of $f (x) = x^{2} - 5 x + 3$ .

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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:

$[- \frac{13}{4}, \infty)$

Step 5.3

Find the domain of $\frac{5 + \sqrt{13 + 4 x}}{2}$ .

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Step 5.3.1

Set the radicand in $\sqrt{13 + 4 x}$ greater than or equal to $0$ to find where the expression is defined.

$13 + 4 x \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Subtract $13$ from both sides of the inequality.

$4 x \geq - 13$

Step 5.3.2.2

Divide each term in $4 x \geq - 13$ by $4$ and simplify.

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Step 5.3.2.2.1

Divide each term in $4 x \geq - 13$ by $4$ .

$\frac{4 x}{4} \geq \frac{- 13}{4}$

Step 5.3.2.2.2

Simplify the left side.

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Step 5.3.2.2.2.1

Cancel the common factor of $4$ .

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Step 5.3.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} \geq \frac{- 13}{4}$

Step 5.3.2.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{- 13}{4}$

Step 5.3.2.2.3

Simplify the right side.

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Step 5.3.2.2.3.1

Move the negative in front of the fraction.

$x \geq - \frac{13}{4}$

Step 5.3.3

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

$[- \frac{13}{4}, \infty)$

Step 5.4

Find the domain of $f (x) = x^{2} - 5 x + 3$ .

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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 $f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$ is the range of $f (x) = x^{2} - 5 x + 3$ and the range of $f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$ is the domain of $f (x) = x^{2} - 5 x + 3$ , then $f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$ is the inverse of $f (x) = x^{2} - 5 x + 3$ .

$f^{- 1} (x) = \frac{5 + \sqrt{13 + 4 x}}{2}, \frac{5 - \sqrt{13 + 4 x}}{2}$

Step 6