Precalculus Examples

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

Step 1

Write $f (x) = 1 + \sqrt{1 + x}$ as an equation.

$y = 1 + \sqrt{1 + x}$

Step 2

Interchange the variables.

$x = 1 + \sqrt{1 + y}$

Step 3

Solve for $y$ .

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Step 3.1

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

$1 + \sqrt{1 + y} = x$

Step 3.2

Subtract $1$ from both sides of the equation.

$\sqrt{1 + y} = x - 1$

Step 3.3

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

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

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 $\sqrt{1 + y}$ as ${(1 + y)}^{\frac{1}{2}}$ .

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

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

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

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Step 3.4.2.1.1.1

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

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

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.

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

Step 3.4.2.1.1.2.2

Rewrite the expression.

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

Step 3.4.2.1.2

Simplify.

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

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Simplify ${(x - 1)}^{2}$ .

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Step 3.4.3.1.1

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$1 + y = (x - 1) (x - 1)$

Step 3.4.3.1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 3.4.3.1.2.1

Apply the distributive property.

$1 + y = x (x - 1) - 1 (x - 1)$

Step 3.4.3.1.2.2

Apply the distributive property.

$1 + y = x \cdot x + x \cdot - 1 - 1 (x - 1)$

Step 3.4.3.1.2.3

Apply the distributive property.

$1 + y = x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1$

Step 3.4.3.1.3

Simplify and combine like terms.

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Step 3.4.3.1.3.1

Simplify each term.

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Step 3.4.3.1.3.1.1

Multiply $x$ by $x$ .

$1 + y = x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1$

Step 3.4.3.1.3.1.2

Move $- 1$ to the left of $x$ .

$1 + y = x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1$

Step 3.4.3.1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$1 + y = x^{2} - x - 1 x - 1 \cdot - 1$

Step 3.4.3.1.3.1.4

Rewrite $- 1 x$ as $- x$ .

$1 + y = x^{2} - x - x - 1 \cdot - 1$

Step 3.4.3.1.3.1.5

Multiply $- 1$ by $- 1$ .

$1 + y = x^{2} - x - x + 1$

Step 3.4.3.1.3.2

Subtract $x$ from $- x$ .

$1 + y = x^{2} - 2 x + 1$

Step 3.5

Move all terms not containing $y$ to the right side of the equation.

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Step 3.5.1

Subtract $1$ from both sides of the equation.

$y = x^{2} - 2 x + 1 - 1$

Step 3.5.2

Combine the opposite terms in $x^{2} - 2 x + 1 - 1$ .

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Step 3.5.2.1

Subtract $1$ from $1$ .

$y = x^{2} - 2 x + 0$

Step 3.5.2.2

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

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Rewrite ${(1 + \sqrt{1 + x})}^{2}$ as $(1 + \sqrt{1 + x}) (1 + \sqrt{1 + x})$ .

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

Step 5.2.3.2

Expand $(1 + \sqrt{1 + x}) (1 + \sqrt{1 + x})$ using the FOIL Method.

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Step 5.2.3.2.1

Apply the distributive property.

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

Step 5.2.3.2.2

Apply the distributive property.

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

Step 5.2.3.2.3

Apply the distributive property.

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

Step 5.2.3.3

Simplify and combine like terms.

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Step 5.2.3.3.1

Simplify each term.

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Step 5.2.3.3.1.1

Multiply $1$ by $1$ .

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

Step 5.2.3.3.1.2

Multiply $\sqrt{1 + x}$ by $1$ .

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

Step 5.2.3.3.1.3

Multiply $\sqrt{1 + x}$ by $1$ .

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

Step 5.2.3.3.1.4

Multiply $\sqrt{1 + x} \sqrt{1 + x}$ .

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Step 5.2.3.3.1.4.1

Raise $\sqrt{1 + x}$ to the power of $1$ .

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

Step 5.2.3.3.1.4.2

Raise $\sqrt{1 + x}$ to the power of $1$ .

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

Step 5.2.3.3.1.4.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.2.3.3.1.4.4

Add $1$ and $1$ .

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

Step 5.2.3.3.1.5

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

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Step 5.2.3.3.1.5.1

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

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

Step 5.2.3.3.1.5.2

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

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

Step 5.2.3.3.1.5.3

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

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

Step 5.2.3.3.1.5.4

Cancel the common factor of $2$ .

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Step 5.2.3.3.1.5.4.1

Cancel the common factor.

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

Step 5.2.3.3.1.5.4.2

Rewrite the expression.

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

Step 5.2.3.3.1.5.5

Simplify.

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

Step 5.2.3.3.2

Add $1$ and $1$ .

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

Step 5.2.3.3.3

Add $\sqrt{1 + x}$ and $\sqrt{1 + x}$ .

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

Step 5.2.3.4

Apply the distributive property.

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

Step 5.2.3.5

Multiply $- 2$ by $1$ .

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

Step 5.2.4

Combine the opposite terms in $2 + 2 \sqrt{1 + x} + x - 2 - 2 \sqrt{1 + x}$ .

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Step 5.2.4.1

Subtract $2$ from $2$ .

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

Step 5.2.4.2

Add $0$ and $2 \sqrt{1 + x}$ .

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

Step 5.2.4.3

Subtract $2 \sqrt{1 + x}$ from $2 \sqrt{1 + x}$ .

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

Step 5.2.4.4

Add $x$ and $0$ .

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

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

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

Step 5.3.3

Remove parentheses.

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

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

Factor using the perfect square rule.

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Step 5.3.4.1.1

Rearrange terms.

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

Step 5.3.4.1.2

Rewrite $1$ as $1^{2}$ .

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

Step 5.3.4.1.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot 1 \cdot x$

Step 5.3.4.1.4

Rewrite the polynomial.

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

Step 5.3.4.1.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 1$ and $b = x$ .

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

Step 5.3.4.2

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

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

Step 5.3.5

Simplify the expression.

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Step 5.3.5.1

Add $1$ and $1$ .

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

Step 5.3.5.2

Reorder $2$ and $- x$ .

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

Step 5.4

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

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