Calculus Examples

$\sqrt{x - 1}$

Step 1

Interchange the variables.

$x = \sqrt{y - 1}$

Step 2

Solve for $y$ .

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Rewrite the equation as $\sqrt{y - 1} = x$ .

$\sqrt{y - 1} = x$

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

${\sqrt{y - 1}}^{2} = x^{2}$

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y - 1 = x^{2}$

Add $1$ to both sides of the equation.

$y = x^{2} + 1$

Step 3

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

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

Step 4

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

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

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

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Set up the composite result function.

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

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

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

Rewrite ${\sqrt{x - 1}}^{2}$ as $x - 1$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine the opposite terms in $x - 1 + 1$ .

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Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

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Set up the composite result function.

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

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

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

Subtract $1$ from $1$ .

$f (x^{2} + 1) = \sqrt{x^{2} + 0}$

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

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

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

$f (x^{2} + 1) = x$

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

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