Trigonometry Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Subtract $2$ from both sides of the equation.

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

Step 3.3

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

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

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 $\sqrt{2 y - 3}$ as ${(2 y - 3)}^{\frac{1}{2}}$ .

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

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Simplify ${({(2 y - 3)}^{\frac{1}{2}})}^{2}$ .

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Step 3.4.2.1.1

Multiply the exponents in ${({(2 y - 3)}^{\frac{1}{2}})}^{2}$ .

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Step 3.4.2.1.1.1

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

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

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.

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

Step 3.4.2.1.1.2.2

Rewrite the expression.

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

Step 3.4.2.1.2

Simplify.

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

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

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

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Step 3.4.3.1.1

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

$2 y - 3 = (x - 2) (x - 2)$

Step 3.4.3.1.2

Expand $(x - 2) (x - 2)$ using the FOIL Method.

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Step 3.4.3.1.2.1

Apply the distributive property.

$2 y - 3 = x (x - 2) - 2 (x - 2)$

Step 3.4.3.1.2.2

Apply the distributive property.

$2 y - 3 = x \cdot x + x \cdot - 2 - 2 (x - 2)$

Step 3.4.3.1.2.3

Apply the distributive property.

$2 y - 3 = x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2$

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 $x$ .

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

Step 3.4.3.1.3.1.2

Move $- 2$ to the left of $x$ .

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

Step 3.4.3.1.3.1.3

Multiply $- 2$ by $- 2$ .

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

Step 3.4.3.1.3.2

Subtract $2 x$ from $- 2 x$ .

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

Step 3.5

Solve for $y$ .

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Step 3.5.1

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

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Step 3.5.1.1

Add $3$ to both sides of the equation.

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

Step 3.5.1.2

Add $4$ and $3$ .

$2 y = x^{2} - 4 x + 7$

Step 3.5.2

Divide each term in $2 y = x^{2} - 4 x + 7$ by $2$ and simplify.

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Step 3.5.2.1

Divide each term in $2 y = x^{2} - 4 x + 7$ by $2$ .

$\frac{2 y}{2} = \frac{x^{2}}{2} + \frac{- 4 x}{2} + \frac{7}{2}$

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $2$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{x^{2}}{2} + \frac{- 4 x}{2} + \frac{7}{2}$

Step 3.5.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{2}}{2} + \frac{- 4 x}{2} + \frac{7}{2}$

Step 3.5.2.3

Simplify the right side.

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Step 3.5.2.3.1

Cancel the common factor of $- 4$ and $2$ .

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Step 3.5.2.3.1.1

Factor $2$ out of $- 4 x$ .

$y = \frac{x^{2}}{2} + \frac{2 (- 2 x)}{2} + \frac{7}{2}$

Step 3.5.2.3.1.2

Cancel the common factors.

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Step 3.5.2.3.1.2.1

Factor $2$ out of $2$ .

$y = \frac{x^{2}}{2} + \frac{2 (- 2 x)}{2 (1)} + \frac{7}{2}$

Step 3.5.2.3.1.2.2

Cancel the common factor.

$y = \frac{x^{2}}{2} + \frac{2 (- 2 x)}{2 \cdot 1} + \frac{7}{2}$

Step 3.5.2.3.1.2.3

Rewrite the expression.

$y = \frac{x^{2}}{2} + \frac{- 2 x}{1} + \frac{7}{2}$

Step 3.5.2.3.1.2.4

Divide $- 2 x$ by $1$ .

$y = \frac{x^{2}}{2} - 2 x + \frac{7}{2}$

Step 4

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

$f^{- 1} (x) = \frac{x^{2}}{2} - 2 x + \frac{7}{2}$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{2}}{2} - 2 x + \frac{7}{2}$ is the inverse of $f (x) = \sqrt{2 x - 3} + 2$ .

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

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(\sqrt{2 x - 3} + 2)}^{2}}{2} - 2 (\sqrt{2 x - 3} + 2) + \frac{7}{2}$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Apply the distributive property.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(\sqrt{2 x - 3} + 2)}^{2}}{2} - 2 \sqrt{2 x - 3} - 2 \cdot 2 + \frac{7}{2}$

Step 5.2.3.2

Multiply $- 2$ by $2$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(\sqrt{2 x - 3} + 2)}^{2}}{2} - 2 \sqrt{2 x - 3} - 4 + \frac{7}{2}$

Step 5.2.4

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

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(\sqrt{2 x - 3} + 2)}^{2}}{2} - 2 \sqrt{2 x - 3} \cdot \frac{2}{2} - 4 + \frac{7}{2}$

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine $- 2 \sqrt{2 x - 3}$ and $\frac{2}{2}$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(\sqrt{2 x - 3} + 2)}^{2}}{2} + \frac{- 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.5.2

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(\sqrt{2 x - 3} + 2)}^{2} - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6

Simplify the numerator.

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Step 5.2.6.1

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

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{(\sqrt{2 x - 3} + 2) (\sqrt{2 x - 3} + 2) - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.2

Expand $(\sqrt{2 x - 3} + 2) (\sqrt{2 x - 3} + 2)$ using the FOIL Method.

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Step 5.2.6.2.1

Apply the distributive property.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{\sqrt{2 x - 3} (\sqrt{2 x - 3} + 2) + 2 (\sqrt{2 x - 3} + 2) - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.2.2

Apply the distributive property.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{\sqrt{2 x - 3} \sqrt{2 x - 3} + \sqrt{2 x - 3} \cdot 2 + 2 (\sqrt{2 x - 3} + 2) - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.2.3

Apply the distributive property.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{\sqrt{2 x - 3} \sqrt{2 x - 3} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3

Simplify and combine like terms.

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Step 5.2.6.3.1

Simplify each term.

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Step 5.2.6.3.1.1

Multiply $\sqrt{2 x - 3} \sqrt{2 x - 3}$ .

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Step 5.2.6.3.1.1.1

Raise $\sqrt{2 x - 3}$ to the power of $1$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{\sqrt{2 x - 3} \sqrt{2 x - 3} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.1.2

Raise $\sqrt{2 x - 3}$ to the power of $1$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{\sqrt{2 x - 3} \sqrt{2 x - 3} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.1.3

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

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{\sqrt{2 x - 3}}^{1 + 1} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.1.4

Add $1$ and $1$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{\sqrt{2 x - 3}}^{2} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.2

Rewrite ${\sqrt{2 x - 3}}^{2}$ as $2 x - 3$ .

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Step 5.2.6.3.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x - 3}$ as ${(2 x - 3)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{({(2 x - 3)}^{\frac{1}{2}})}^{2} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.2.2

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

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(2 x - 3)}^{\frac{1}{2} \cdot 2} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.2.3

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

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(2 x - 3)}^{\frac{2}{2}} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.2.4

Cancel the common factor of $2$ .

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Step 5.2.6.3.1.2.4.1

Cancel the common factor.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{{(2 x - 3)}^{\frac{2}{2}} + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.2.4.2

Rewrite the expression.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{(2 x - 3) + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.2.5

Simplify.

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x - 3 + \sqrt{2 x - 3} \cdot 2 + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.3

Move $2$ to the left of $\sqrt{2 x - 3}$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x - 3 + 2 \cdot \sqrt{2 x - 3} + 2 \sqrt{2 x - 3} + 2 \cdot 2 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.1.4

Multiply $2$ by $2$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x - 3 + 2 \sqrt{2 x - 3} + 2 \sqrt{2 x - 3} + 4 - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.2

Add $- 3$ and $4$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x + 1 + 2 \sqrt{2 x - 3} + 2 \sqrt{2 x - 3} - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.3.3

Add $2 \sqrt{2 x - 3}$ and $2 \sqrt{2 x - 3}$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x + 1 + 4 \sqrt{2 x - 3} - 2 \sqrt{2 x - 3} \cdot 2}{2} - 4 + \frac{7}{2}$

Step 5.2.6.4

Multiply $2$ by $- 2$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x + 1 + 4 \sqrt{2 x - 3} - 4 \sqrt{2 x - 3}}{2} - 4 + \frac{7}{2}$

Step 5.2.6.5

Subtract $4 \sqrt{2 x - 3}$ from $4 \sqrt{2 x - 3}$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x + 1 + 0}{2} - 4 + \frac{7}{2}$

Step 5.2.6.6

Add $2 x + 1$ and $0$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = \frac{2 x + 1}{2} - 4 + \frac{7}{2}$

Step 5.2.7

Simplify terms.

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Step 5.2.7.1

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt{2 x - 3} + 2) = - 4 + \frac{2 x + 1 + 7}{2}$

Step 5.2.7.2

Add $1$ and $7$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = - 4 + \frac{2 x + 8}{2}$

Step 5.2.7.3

Cancel the common factor of $2 x + 8$ and $2$ .

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Step 5.2.7.3.1

Factor $2$ out of $2 x$ .

$f^{- 1} (\sqrt{2 x - 3} + 2) = - 4 + \frac{2 (x) + 8}{2}$

Step 5.2.7.3.2

Factor $2$ out of $8$ .

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

Step 5.2.7.3.3

Factor $2$ out of $2 (x) + 2 (4)$ .

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

Step 5.2.7.3.4

Cancel the common factors.

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Step 5.2.7.3.4.1

Factor $2$ out of $2$ .

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

Step 5.2.7.3.4.2

Cancel the common factor.

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

Step 5.2.7.3.4.3

Rewrite the expression.

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

Step 5.2.7.3.4.4

Divide $x + 4$ by $1$ .

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

Step 5.2.7.4

Combine the opposite terms in $- 4 + x + 4$ .

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Step 5.2.7.4.1

Add $- 4$ and $4$ .

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

Step 5.2.7.4.2

Add $x$ and $0$ .

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

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 $f (\frac{x^{2}}{2} - 2 x + \frac{7}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{2 (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) - 3} + 2$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{2 (\frac{x^{2}}{2}) + 2 (- 2 x) + 2 (\frac{7}{2}) - 3} + 2$

Step 5.3.3.2

Simplify.

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Step 5.3.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.3.2.1.1

Cancel the common factor.

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{2 (\frac{x^{2}}{2}) + 2 (- 2 x) + 2 (\frac{7}{2}) - 3} + 2$

Step 5.3.3.2.1.2

Rewrite the expression.

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} + 2 (- 2 x) + 2 (\frac{7}{2}) - 3} + 2$

Step 5.3.3.2.2

Multiply $- 2$ by $2$ .

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} - 4 x + 2 (\frac{7}{2}) - 3} + 2$

Step 5.3.3.2.3

Cancel the common factor of $2$ .

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Step 5.3.3.2.3.1

Cancel the common factor.

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} - 4 x + 2 (\frac{7}{2}) - 3} + 2$

Step 5.3.3.2.3.2

Rewrite the expression.

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} - 4 x + 7 - 3} + 2$

Step 5.3.3.3

Subtract $3$ from $7$ .

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} - 4 x + 4} + 2$

Step 5.3.3.4

Factor using the perfect square rule.

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Step 5.3.3.4.1

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

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} - 4 x + 2^{2}} + 2$

Step 5.3.3.4.2

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

$4 x = 2 \cdot x \cdot 2$

Step 5.3.3.4.3

Rewrite the polynomial.

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{x^{2} - 2 \cdot x \cdot 2 + 2^{2}} + 2$

Step 5.3.3.4.4

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

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = \sqrt{{(x - 2)}^{2}} + 2$

Step 5.3.3.5

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

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = x - 2 + 2$

Step 5.3.4

Combine the opposite terms in $x - 2 + 2$ .

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Step 5.3.4.1

Add $- 2$ and $2$ .

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (\frac{x^{2}}{2} - 2 x + \frac{7}{2}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{2} - 2 x + \frac{7}{2}$ is the inverse of $f (x) = \sqrt{2 x - 3} + 2$ .

$f^{- 1} (x) = \frac{x^{2}}{2} - 2 x + \frac{7}{2}$