Precalculus Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

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

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

Step 2.2

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

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

Step 2.3

Simplify the left side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

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

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

Step 2.3.1.1.2

Cancel the common factor of $2$ .

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Step 2.3.1.1.2.1

Cancel the common factor.

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

Step 2.3.1.1.2.2

Rewrite the expression.

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

Step 2.3.1.2

Simplify.

$y + 5 = x^{2}$

Step 2.4

Subtract $5$ from both sides of the equation.

$y = x^{2} - 5$

Step 3

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

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

Step 4

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

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

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

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Multiply the exponents in ${({(x + 5)}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.3.1.1

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

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

Step 4.2.3.1.2

Cancel the common factor of $2$ .

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Step 4.2.3.1.2.1

Cancel the common factor.

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

Step 4.2.3.1.2.2

Rewrite the expression.

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

Step 4.2.3.2

Simplify.

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

Step 4.2.4

Combine the opposite terms in $x + 5 - 5$ .

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Step 4.2.4.1

Subtract $5$ from $5$ .

$f^{- 1} ({(x + 5)}^{\frac{1}{2}}) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

Combine the opposite terms in $(x^{2} - 5) + 5$ .

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Step 4.3.3.1

Add $- 5$ and $5$ .

$f (x^{2} - 5) = {(x^{2} + 0)}^{\frac{1}{2}}$

Step 4.3.3.2

Add $x^{2}$ and $0$ .

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

Step 4.3.4

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

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Step 4.3.4.1

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

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

Step 4.3.4.2

Cancel the common factor of $2$ .

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Step 4.3.4.2.1

Cancel the common factor.

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

Step 4.3.4.2.2

Rewrite the expression.

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

Step 4.4

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

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