Trigonometry Examples

$f (x) = \sqrt{5 x - 20}$

Step 1

Write $f (x) = \sqrt{5 x - 20}$ as an equation.

$y = \sqrt{5 x - 20}$

Step 2

Interchange the variables.

$x = \sqrt{5 y - 20}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\sqrt{5 y - 20} = x$ .

$\sqrt{5 y - 20} = x$

Step 3.2

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{5 y - 20}}^{2} = x^{2}$

Step 3.3

Simplify each side of the equation.

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Step 3.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 y - 20}$ as ${(5 y - 20)}^{\frac{1}{2}}$ .

${({(5 y - 20)}^{\frac{1}{2}})}^{2} = x^{2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Simplify ${({(5 y - 20)}^{\frac{1}{2}})}^{2}$ .

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Step 3.3.2.1.1

Multiply the exponents in ${({(5 y - 20)}^{\frac{1}{2}})}^{2}$ .

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Step 3.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(5 y - 20)}^{\frac{1}{2} \cdot 2} = x^{2}$

Step 3.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.3.2.1.1.2.1

Cancel the common factor.

${(5 y - 20)}^{\frac{1}{2} \cdot 2} = x^{2}$

Step 3.3.2.1.1.2.2

Rewrite the expression.

${(5 y - 20)}^{1} = x^{2}$

Step 3.3.2.1.2

Simplify.

$5 y - 20 = x^{2}$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Add $20$ to both sides of the equation.

$5 y = x^{2} + 20$

Step 3.4.2

Divide each term in $5 y = x^{2} + 20$ by $5$ and simplify.

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Step 3.4.2.1

Divide each term in $5 y = x^{2} + 20$ by $5$ .

$\frac{5 y}{5} = \frac{x^{2}}{5} + \frac{20}{5}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $5$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{5 y}{5} = \frac{x^{2}}{5} + \frac{20}{5}$

Step 3.4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{2}}{5} + \frac{20}{5}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

Divide $20$ by $5$ .

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

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = \frac{x^{2}}{5} + 4$ is the inverse of $f (x) = \sqrt{5 x - 20}$ .

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Step 5.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 5.2

Evaluate $f^{- 1} (f (x))$ .

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Step 5.2.1

Set up the composite result function.

$f^{- 1} (f (x))$

Step 5.2.2

Evaluate $f^{- 1} (\sqrt{5 x - 20})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{(\sqrt{5 x - 20})}^{2}}{5} + 4$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

Factor $5$ out of $5 x - 20$ .

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Step 5.2.3.1.1.1

Factor $5$ out of $5 x$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{\sqrt{5 (x) - 20}}^{2}}{5} + 4$

Step 5.2.3.1.1.2

Factor $5$ out of $- 20$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{\sqrt{5 x + 5 \cdot - 4}}^{2}}{5} + 4$

Step 5.2.3.1.1.3

Factor $5$ out of $5 x + 5 \cdot - 4$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{\sqrt{5 (x - 4)}}^{2}}{5} + 4$

Step 5.2.3.1.2

Rewrite ${\sqrt{5 (x - 4)}}^{2}$ as $5 (x - 4)$ .

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Step 5.2.3.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 (x - 4)}$ as ${(5 (x - 4))}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{({(5 (x - 4))}^{\frac{1}{2}})}^{2}}{5} + 4$

Step 5.2.3.1.2.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{(5 (x - 4))}^{\frac{1}{2} \cdot 2}}{5} + 4$

Step 5.2.3.1.2.3

Combine $\frac{1}{2}$ and $2$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{(5 (x - 4))}^{\frac{2}{2}}}{5} + 4$

Step 5.2.3.1.2.4

Cancel the common factor of $2$ .

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Step 5.2.3.1.2.4.1

Cancel the common factor.

$f^{- 1} (\sqrt{5 x - 20}) = \frac{{(5 (x - 4))}^{\frac{2}{2}}}{5} + 4$

Step 5.2.3.1.2.4.2

Rewrite the expression.

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 (x - 4)}{5} + 4$

Step 5.2.3.1.2.5

Simplify.

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 (x - 4)}{5} + 4$

Step 5.2.3.1.3

Apply the distributive property.

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 x + 5 \cdot - 4}{5} + 4$

Step 5.2.3.1.4

Multiply $5$ by $- 4$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 x - 20}{5} + 4$

Step 5.2.3.1.5

Factor $5$ out of $5 x - 20$ .

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Step 5.2.3.1.5.1

Factor $5$ out of $5 x$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 (x) - 20}{5} + 4$

Step 5.2.3.1.5.2

Factor $5$ out of $- 20$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 x + 5 \cdot - 4}{5} + 4$

Step 5.2.3.1.5.3

Factor $5$ out of $5 x + 5 \cdot - 4$ .

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 (x - 4)}{5} + 4$

Step 5.2.3.2

Cancel the common factor of $5$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (\sqrt{5 x - 20}) = \frac{5 (x - 4)}{5} + 4$

Step 5.2.3.2.2

Divide $x - 4$ by $1$ .

$f^{- 1} (\sqrt{5 x - 20}) = x - 4 + 4$

Step 5.2.4

Combine the opposite terms in $x - 4 + 4$ .

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Step 5.2.4.1

Add $- 4$ and $4$ .

$f^{- 1} (\sqrt{5 x - 20}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\sqrt{5 x - 20}) = x$

Step 5.3

Evaluate $f (f^{- 1} (x))$ .

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Step 5.3.1

Set up the composite result function.

$f (f^{- 1} (x))$

Step 5.3.2

Evaluate $f (\frac{x^{2}}{5} + 4)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{5} + 4) = \sqrt{5 (\frac{x^{2}}{5} + 4) - 20}$

Step 5.3.3

Factor $5$ out of $5 (\frac{x^{2}}{5} + 4) - 20$ .

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Step 5.3.3.1

Factor $5$ out of $- 20$ .

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

Step 5.3.3.2

Factor $5$ out of $5 (\frac{x^{2}}{5} + 4) + 5 \cdot - 4$ .

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

Step 5.3.4

Simplify by subtracting numbers.

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Step 5.3.4.1

Subtract $4$ from $4$ .

$f (\frac{x^{2}}{5} + 4) = \sqrt{5 (\frac{x^{2}}{5} + 0)}$

Step 5.3.4.2

Add $\frac{x^{2}}{5}$ and $0$ .

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

Step 5.3.5

Combine $5$ and $\frac{x^{2}}{5}$ .

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

Step 5.3.6

Reduce the expression by cancelling the common factors.

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Step 5.3.6.1

Reduce the expression $\frac{5 x^{2}}{5}$ by cancelling the common factors.

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Step 5.3.6.1.1

Cancel the common factor.

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

Step 5.3.6.1.2

Rewrite the expression.

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

Step 5.3.6.2

Divide $x^{2}$ by $1$ .

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

Step 5.3.7

Pull terms out from under the radical, assuming positive real numbers.

$f (\frac{x^{2}}{5} + 4) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{5} + 4$ is the inverse of $f (x) = \sqrt{5 x - 20}$ .

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