Trigonometry Examples

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

Step 1

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

$y = \sqrt{w - 1}$

Step 2

Interchange the variables.

$w = \sqrt{y - 1}$

Step 3

Solve for $y$ .

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Step 3.1

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

$\sqrt{y - 1} = w$

Step 3.2

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

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

Step 3.3

Simplify each side of the equation.

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

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

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Step 3.3.2.1.1.1

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

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

Step 3.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.3.2.1.1.2.1

Cancel the common factor.

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

Step 3.3.2.1.1.2.2

Rewrite the expression.

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

Step 3.3.2.1.2

Simplify.

$y - 1 = w^{2}$

Step 3.4

Add $1$ to both sides of the equation.

$y = w^{2} + 1$

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

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

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Step 5.2.3.1

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

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

Step 5.2.3.2

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

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

Step 5.2.3.3

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

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

Step 5.2.3.4

Cancel the common factor of $2$ .

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Step 5.2.3.4.1

Cancel the common factor.

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

Step 5.2.3.4.2

Rewrite the expression.

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

Step 5.2.3.5

Simplify.

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

Step 5.2.4

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

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Step 5.2.4.1

Add $- 1$ and $1$ .

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

Step 5.2.4.2

Add $w$ and $0$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

Subtract $1$ from $1$ .

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

Step 5.3.4

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

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

Step 5.3.5

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

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

Step 5.4

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

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