Precalculus Examples

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

Step 1

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

$y = 5 \sqrt{x}$

Step 2

Interchange the variables.

$x = 5 \sqrt{y}$

Step 3

Solve for $y$ .

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Step 3.1

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

$5 \sqrt{y} = x$

Step 3.2

Divide each term in $5 \sqrt{y} = x$ by $5$ and simplify.

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Step 3.2.1

Divide each term in $5 \sqrt{y} = x$ by $5$ .

$\frac{5 \sqrt{y}}{5} = \frac{x}{5}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $5$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{5 \sqrt{y}}{5} = \frac{x}{5}$

Step 3.2.2.1.2

Divide $\sqrt{y}$ by $1$ .

$\sqrt{y} = \frac{x}{5}$

Step 3.3

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

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

Step 3.4

Simplify each side of the equation.

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Step 3.4.1

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

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

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

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

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Step 3.4.2.1.1.1

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

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

Step 3.4.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.4.2.1.1.2.1

Cancel the common factor.

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

Step 3.4.2.1.1.2.2

Rewrite the expression.

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

Step 3.4.2.1.2

Simplify.

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

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Simplify ${(\frac{x}{5})}^{2}$ .

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Step 3.4.3.1.1

Apply the product rule to $\frac{x}{5}$ .

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

Step 3.4.3.1.2

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

$y = \frac{x^{2}}{25}$

Step 4

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

$f^{- 1} (x) = \frac{x^{2}}{25}$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{2}}{25}$ is the inverse of $f (x) = 5 \sqrt{x}$ .

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

$f^{- 1} (5 \sqrt{x}) = \frac{{(5 \sqrt{x})}^{2}}{25}$

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Apply the product rule to $5 \sqrt{x}$ .

$f^{- 1} (5 \sqrt{x}) = \frac{5^{2} {\sqrt{x}}^{2}}{25}$

Step 5.2.3.2

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

$f^{- 1} (5 \sqrt{x}) = \frac{25 {\sqrt{x}}^{2}}{25}$

Step 5.2.3.3

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

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Step 5.2.3.3.1

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

$f^{- 1} (5 \sqrt{x}) = \frac{25 {(x^{\frac{1}{2}})}^{2}}{25}$

Step 5.2.3.3.2

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

$f^{- 1} (5 \sqrt{x}) = \frac{25 x^{\frac{1}{2} \cdot 2}}{25}$

Step 5.2.3.3.3

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

$f^{- 1} (5 \sqrt{x}) = \frac{25 x^{\frac{2}{2}}}{25}$

Step 5.2.3.3.4

Cancel the common factor of $2$ .

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Step 5.2.3.3.4.1

Cancel the common factor.

$f^{- 1} (5 \sqrt{x}) = \frac{25 x^{\frac{2}{2}}}{25}$

Step 5.2.3.3.4.2

Rewrite the expression.

$f^{- 1} (5 \sqrt{x}) = \frac{25 x}{25}$

Step 5.2.3.3.5

Simplify.

$f^{- 1} (5 \sqrt{x}) = \frac{25 x}{25}$

Step 5.2.4

Cancel the common factor of $25$ .

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Step 5.2.4.1

Cancel the common factor.

$f^{- 1} (5 \sqrt{x}) = \frac{25 x}{25}$

Step 5.2.4.2

Divide $x$ by $1$ .

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

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 $f (\frac{x^{2}}{25})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{25}) = 5 \sqrt{\frac{x^{2}}{25}}$

Step 5.3.3

Remove parentheses.

$f (\frac{x^{2}}{25}) = 5 \sqrt{\frac{x^{2}}{25}}$

Step 5.3.4

Rewrite $25$ as $5^{2}$ .

$f (\frac{x^{2}}{25}) = 5 \sqrt{\frac{x^{2}}{5^{2}}}$

Step 5.3.5

Rewrite $\frac{x^{2}}{5^{2}}$ as ${(\frac{x}{5})}^{2}$ .

$f (\frac{x^{2}}{25}) = 5 \sqrt{{(\frac{x}{5})}^{2}}$

Step 5.3.6

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

$f (\frac{x^{2}}{25}) = 5 (\frac{x}{5})$

Step 5.3.7

Cancel the common factor of $5$ .

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Step 5.3.7.1

Cancel the common factor.

$f (\frac{x^{2}}{25}) = 5 (\frac{x}{5})$

Step 5.3.7.2

Rewrite the expression.

$f (\frac{x^{2}}{25}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{25}$ is the inverse of $f (x) = 5 \sqrt{x}$ .

$f^{- 1} (x) = \frac{x^{2}}{25}$