Algebra Examples

$f (x) = \frac{1}{5} {(x - 2)}^{2} + 3$

Step 1

Write $f (x) = \frac{1}{5} {(x - 2)}^{2} + 3$ as an equation.

$y = \frac{1}{5} {(x - 2)}^{2} + 3$

Step 2

Interchange the variables.

$x = \frac{1}{5} \cdot {(y - 2)}^{2} + 3$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{1}{5} \cdot {(y - 2)}^{2} + 3 = x$ .

$\frac{1}{5} \cdot {(y - 2)}^{2} + 3 = x$

Step 3.2

Subtract $3$ from both sides of the equation.

$\frac{1}{5} \cdot {(y - 2)}^{2} = x - 3$

Step 3.3

Combine $\frac{1}{5}$ and ${(y - 2)}^{2}$ .

$\frac{{(y - 2)}^{2}}{5} = x - 3$

Step 3.4

Multiply both sides of the equation by $5$ .

$5 \frac{{(y - 2)}^{2}}{5} = 5 (x - 3)$

Step 3.5

Simplify both sides of the equation.

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Step 3.5.1

Simplify the left side.

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Step 3.5.1.1

Cancel the common factor of $5$ .

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Step 3.5.1.1.1

Cancel the common factor.

$5 \frac{{(y - 2)}^{2}}{5} = 5 (x - 3)$

Step 3.5.1.1.2

Rewrite the expression.

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

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $5 (x - 3)$ .

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Step 3.5.2.1.1

Apply the distributive property.

${(y - 2)}^{2} = 5 x + 5 \cdot - 3$

Step 3.5.2.1.2

Multiply $5$ by $- 3$ .

${(y - 2)}^{2} = 5 x - 15$

Step 3.6

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

$y - 2 = \pm \sqrt{5 x - 15}$

Step 3.7

Factor $5$ out of $5 x - 15$ .

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Step 3.7.1

Factor $5$ out of $5 x$ .

$y - 2 = \pm \sqrt{5 (x) - 15}$

Step 3.7.2

Factor $5$ out of $- 15$ .

$y - 2 = \pm \sqrt{5 x + 5 \cdot - 3}$

Step 3.7.3

Factor $5$ out of $5 x + 5 \cdot - 3$ .

$y - 2 = \pm \sqrt{5 (x - 3)}$

Step 3.8

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

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Step 3.8.1

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

$y - 2 = \sqrt{5 (x - 3)}$

Step 3.8.2

Add $2$ to both sides of the equation.

$y = \sqrt{5 (x - 3)} + 2$

Step 3.8.3

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

$y - 2 = - \sqrt{5 (x - 3)}$

Step 3.8.4

Add $2$ to both sides of the equation.

$y = - \sqrt{5 (x - 3)} + 2$

Step 3.8.5

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

$y = \sqrt{5 (x - 3)} + 2$

$y = - \sqrt{5 (x - 3)} + 2$

$y = \sqrt{5 (x - 3)} + 2$

$y = - \sqrt{5 (x - 3)} + 2$

$y = \sqrt{5 (x - 3)} + 2$

$y = - \sqrt{5 (x - 3)} + 2$

Step 4

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

$f^{- 1} (x) = \sqrt{5 (x - 3)} + 2, - \sqrt{5 (x - 3)} + 2$

Step 5

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

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

Step 5.2

Find the range of $f (x) = \frac{1}{5} \cdot {(x - 2)}^{2} + 3$ .

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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:

$[3, \infty)$

Step 5.3

Find the domain of $\sqrt{5 (x - 3)} + 2$ .

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Step 5.3.1

Set the radicand in $\sqrt{5 (x - 3)}$ greater than or equal to $0$ to find where the expression is defined.

$5 (x - 3) \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Divide each term in $5 (x - 3) \geq 0$ by $5$ and simplify.

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Step 5.3.2.1.1

Divide each term in $5 (x - 3) \geq 0$ by $5$ .

$\frac{5 (x - 3)}{5} \geq \frac{0}{5}$

Step 5.3.2.1.2

Simplify the left side.

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Step 5.3.2.1.2.1

Cancel the common factor of $5$ .

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Step 5.3.2.1.2.1.1

Cancel the common factor.

$\frac{5 (x - 3)}{5} \geq \frac{0}{5}$

Step 5.3.2.1.2.1.2

Divide $x - 3$ by $1$ .

$x - 3 \geq \frac{0}{5}$

Step 5.3.2.1.3

Simplify the right side.

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Step 5.3.2.1.3.1

Divide $0$ by $5$ .

$x - 3 \geq 0$

Step 5.3.2.2

Add $3$ to both sides of the inequality.

$x \geq 3$

Step 5.3.3

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

$[3, \infty)$

Step 5.4

Find the domain of $f (x) = \frac{1}{5} \cdot {(x - 2)}^{2} + 3$ .

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

$f^{- 1} (x) = \sqrt{5 (x - 3)} + 2, - \sqrt{5 (x - 3)} + 2$

Step 6