Precalculus Examples

$f (x) = 2 \sqrt{x} + 3$

Step 1

Write $f (x) = 2 \sqrt{x} + 3$ as an equation.

$y = 2 \sqrt{x} + 3$

Step 2

Interchange the variables.

$x = 2 \sqrt{y} + 3$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 \sqrt{y} + 3 = x$ .

$2 \sqrt{y} + 3 = x$

Step 3.2

Subtract $3$ from both sides of the equation.

$2 \sqrt{y} = x - 3$

Step 3.3

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

${(2 \sqrt{y})}^{2} = {(x - 3)}^{2}$

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 $y^{\frac{1}{2}}$ .

${(2 y^{\frac{1}{2}})}^{2} = {(x - 3)}^{2}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

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

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Step 3.4.2.1.1

Apply the product rule to $2 y^{\frac{1}{2}}$ .

$2^{2} {(y^{\frac{1}{2}})}^{2} = {(x - 3)}^{2}$

Step 3.4.2.1.2

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

$4 {(y^{\frac{1}{2}})}^{2} = {(x - 3)}^{2}$

Step 3.4.2.1.3

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

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Step 3.4.2.1.3.1

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

$4 y^{\frac{1}{2} \cdot 2} = {(x - 3)}^{2}$

Step 3.4.2.1.3.2

Cancel the common factor of $2$ .

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Step 3.4.2.1.3.2.1

Cancel the common factor.

$4 y^{\frac{1}{2} \cdot 2} = {(x - 3)}^{2}$

Step 3.4.2.1.3.2.2

Rewrite the expression.

$4 y^{1} = {(x - 3)}^{2}$

Step 3.4.2.1.4

Simplify.

$4 y = {(x - 3)}^{2}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Simplify ${(x - 3)}^{2}$ .

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Step 3.4.3.1.1

Rewrite ${(x - 3)}^{2}$ as $(x - 3) (x - 3)$ .

$4 y = (x - 3) (x - 3)$

Step 3.4.3.1.2

Expand $(x - 3) (x - 3)$ using the FOIL Method.

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Step 3.4.3.1.2.1

Apply the distributive property.

$4 y = x (x - 3) - 3 (x - 3)$

Step 3.4.3.1.2.2

Apply the distributive property.

$4 y = x \cdot x + x \cdot - 3 - 3 (x - 3)$

Step 3.4.3.1.2.3

Apply the distributive property.

$4 y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

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 $x$ .

$4 y = x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3$

Step 3.4.3.1.3.1.2

Move $- 3$ to the left of $x$ .

$4 y = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3$

Step 3.4.3.1.3.1.3

Multiply $- 3$ by $- 3$ .

$4 y = x^{2} - 3 x - 3 x + 9$

Step 3.4.3.1.3.2

Subtract $3 x$ from $- 3 x$ .

$4 y = x^{2} - 6 x + 9$

Step 3.5

Divide each term in $4 y = x^{2} - 6 x + 9$ by $4$ and simplify.

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Step 3.5.1

Divide each term in $4 y = x^{2} - 6 x + 9$ by $4$ .

$\frac{4 y}{4} = \frac{x^{2}}{4} + \frac{- 6 x}{4} + \frac{9}{4}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $4$ .

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Step 3.5.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{x^{2}}{4} + \frac{- 6 x}{4} + \frac{9}{4}$

Step 3.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{2}}{4} + \frac{- 6 x}{4} + \frac{9}{4}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Simplify each term.

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Step 3.5.3.1.1

Cancel the common factor of $- 6$ and $4$ .

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Step 3.5.3.1.1.1

Factor $2$ out of $- 6 x$ .

$y = \frac{x^{2}}{4} + \frac{2 (- 3 x)}{4} + \frac{9}{4}$

Step 3.5.3.1.1.2

Cancel the common factors.

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Step 3.5.3.1.1.2.1

Factor $2$ out of $4$ .

$y = \frac{x^{2}}{4} + \frac{2 (- 3 x)}{2 (2)} + \frac{9}{4}$

Step 3.5.3.1.1.2.2

Cancel the common factor.

$y = \frac{x^{2}}{4} + \frac{2 (- 3 x)}{2 \cdot 2} + \frac{9}{4}$

Step 3.5.3.1.1.2.3

Rewrite the expression.

$y = \frac{x^{2}}{4} + \frac{- 3 x}{2} + \frac{9}{4}$

Step 3.5.3.1.2

Move the negative in front of the fraction.

$y = \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}$

Step 4

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

$f^{- 1} (x) = \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}$ is the inverse of $f (x) = 2 \sqrt{x} + 3$ .

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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 $f^{- 1} (2 \sqrt{x} + 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2}}{4} - \frac{3 (2 \sqrt{x} + 3)}{2} + \frac{9}{4}$

Step 5.2.3

To write $- \frac{3 (2 \sqrt{x} + 3)}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2}}{4} - \frac{3 (2 \sqrt{x} + 3)}{2} \cdot \frac{2}{2} + \frac{9}{4}$

Step 5.2.4

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.2.4.1

Multiply $\frac{3 (2 \sqrt{x} + 3)}{2}$ by $\frac{2}{2}$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2}}{4} - \frac{3 (2 \sqrt{x} + 3) \cdot 2}{2 \cdot 2} + \frac{9}{4}$

Step 5.2.4.2

Multiply $2$ by $2$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2}}{4} - \frac{3 (2 \sqrt{x} + 3) \cdot 2}{4} + \frac{9}{4}$

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the numerators over the common denominator.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} - 3 (2 \sqrt{x} + 3) \cdot 2}{4} + \frac{9}{4}$

Step 5.2.5.2

Combine the numerators over the common denominator.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} - 3 (2 \sqrt{x} + 3) \cdot 2 + 9}{4}$

Step 5.2.5.3

Simplify each term.

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Step 5.2.5.3.1

Apply the distributive property.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} + (- 3 (2 \sqrt{x}) - 3 \cdot 3) \cdot 2 + 9}{4}$

Step 5.2.5.3.2

Multiply $2$ by $- 3$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} + (- 6 \sqrt{x} - 3 \cdot 3) \cdot 2 + 9}{4}$

Step 5.2.5.3.3

Multiply $- 3$ by $3$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} + (- 6 \sqrt{x} - 9) \cdot 2 + 9}{4}$

Step 5.2.5.3.4

Apply the distributive property.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} - 6 \sqrt{x} \cdot 2 - 9 \cdot 2 + 9}{4}$

Step 5.2.5.3.5

Multiply $2$ by $- 6$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} - 12 \sqrt{x} - 9 \cdot 2 + 9}{4}$

Step 5.2.5.3.6

Multiply $- 9$ by $2$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} - 12 \sqrt{x} - 18 + 9}{4}$

Step 5.2.5.4

Add $- 18$ and $9$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{{(2 \sqrt{x} + 3)}^{2} - 12 \sqrt{x} - 9}{4}$

Step 5.2.6

Simplify the numerator.

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Step 5.2.6.1

Let $u = \sqrt{x}$ . Substitute $u$ for all occurrences of $\sqrt{x}$ .

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Step 5.2.6.1.1

Subtract $12 u$ from $12 u$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 u^{2} + 0 + 9 - 9}{4}$

Step 5.2.6.1.2

Add $4 u^{2}$ and $0$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 u^{2} + 9 - 9}{4}$

Step 5.2.6.1.3

Subtract $9$ from $9$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 u^{2} + 0}{4}$

Step 5.2.6.1.4

Add $4 u^{2}$ and $0$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 u^{2}}{4}$

Step 5.2.6.2

Replace all occurrences of $u$ with $\sqrt{x}$ .

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 {\sqrt{x}}^{2}}{4}$

Step 5.2.6.3

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

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Step 5.2.6.3.1

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

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 {(x^{\frac{1}{2}})}^{2}}{4}$

Step 5.2.6.3.2

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

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 x^{\frac{1}{2} \cdot 2}}{4}$

Step 5.2.6.3.3

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

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 x^{\frac{2}{2}}}{4}$

Step 5.2.6.3.4

Cancel the common factor of $2$ .

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Step 5.2.6.3.4.1

Cancel the common factor.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 x^{\frac{2}{2}}}{4}$

Step 5.2.6.3.4.2

Rewrite the expression.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 x}{4}$

Step 5.2.6.3.5

Simplify.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 x}{4}$

Step 5.2.7

Cancel the common factor of $4$ .

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Step 5.2.7.1

Cancel the common factor.

$f^{- 1} (2 \sqrt{x} + 3) = \frac{4 x}{4}$

Step 5.2.7.2

Divide $x$ by $1$ .

$f^{- 1} (2 \sqrt{x} + 3) = x$

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 $f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}} + 3$

Step 5.3.3

Remove parentheses.

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}} + 3$

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

Rewrite $4$ as $2^{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{\frac{x^{2}}{2^{2}} - \frac{3 x}{2} + \frac{9}{4}} + 3$

Step 5.3.4.1.2

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

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x}{2})}^{2} - \frac{3 x}{2} + \frac{9}{4}} + 3$

Step 5.3.4.1.3

Rewrite $9$ as $3^{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x}{2})}^{2} - \frac{3 x}{2} + \frac{3^{2}}{4}} + 3$

Step 5.3.4.1.4

Rewrite $4$ as $2^{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x}{2})}^{2} - \frac{3 x}{2} + \frac{3^{2}}{2^{2}}} + 3$

Step 5.3.4.1.5

Rewrite $\frac{3^{2}}{2^{2}}$ as ${(\frac{3}{2})}^{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x}{2})}^{2} - \frac{3 x}{2} + {(\frac{3}{2})}^{2}} + 3$

Step 5.3.4.1.6

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

$\frac{3 x}{2} = 2 \cdot \frac{x}{2} \cdot \frac{3}{2}$

Step 5.3.4.1.7

Rewrite the polynomial.

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x}{2})}^{2} - 2 \cdot \frac{x}{2} \cdot \frac{3}{2} + {(\frac{3}{2})}^{2}} + 3$

Step 5.3.4.1.8

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = \frac{x}{2}$ and $b = \frac{3}{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x}{2} - \frac{3}{2})}^{2}} + 3$

Step 5.3.4.2

Combine the numerators over the common denominator.

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x - 3}{2})}^{2}} + 3$

Step 5.3.4.3

Apply the product rule to $\frac{x - 3}{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{\frac{{(x - 3)}^{2}}{2^{2}}} + 3$

Step 5.3.4.4

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

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{\frac{{(x - 3)}^{2}}{4}} + 3$

Step 5.3.4.5

Rewrite $4$ as $2^{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{\frac{{(x - 3)}^{2}}{2^{2}}} + 3$

Step 5.3.4.6

Rewrite $\frac{{(x - 3)}^{2}}{2^{2}}$ as ${(\frac{x - 3}{2})}^{2}$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 \sqrt{{(\frac{x - 3}{2})}^{2}} + 3$

Step 5.3.4.7

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

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 (\frac{x - 3}{2}) + 3$

Step 5.3.4.8

Cancel the common factor of $2$ .

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Step 5.3.4.8.1

Cancel the common factor.

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = 2 (\frac{x - 3}{2}) + 3$

Step 5.3.4.8.2

Rewrite the expression.

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = x - 3 + 3$

Step 5.3.5

Combine the opposite terms in $x - 3 + 3$ .

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Step 5.3.5.1

Add $- 3$ and $3$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (\frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}$ is the inverse of $f (x) = 2 \sqrt{x} + 3$ .

$f^{- 1} (x) = \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{9}{4}$