Finite Math Examples

f (x) = sin (\sqrt{e^{x} + 1})

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

Step 1

Write

f (x) = sin (\sqrt{e^{x} + 1})

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

y = sin (\sqrt{e^{x} + 1})

$y = \sin (\sqrt{e^{x} + 1})$

Step 2

Interchange the variables.

x = sin (\sqrt{e^{y} + 1})

$x = \sin (\sqrt{e^{y} + 1})$

Step 3

Solve for

y

$y$ .

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Step 3.1

Rewrite the equation as

sin (\sqrt{e^{y} + 1}) = x

$\sin (\sqrt{e^{y} + 1}) = x$ .

sin (\sqrt{e^{y} + 1}) = x

$\sin (\sqrt{e^{y} + 1}) = x$

Step 3.2

Substitute

u

$u$ for

\sqrt{e^{y} + 1}

$\sqrt{e^{y} + 1}$ .

sin (u) = x

$\sin (u) = x$

Step 3.3

Take the inverse sine of both sides of the equation to extract

u

$u$ from inside the sine.

u = arcsin (x)

$u = \arcsin (x)$

Step 3.4

Substitute

\sqrt{e^{y} + 1}

$\sqrt{e^{y} + 1}$ for

u

$u$ and solve

\sqrt{e^{y} + 1} = arcsin (x)

$\sqrt{e^{y} + 1} = \arcsin (x)$

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Step 3.4.1

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

{\sqrt{e^{y} + 1}}^{2} = {arcsin (x)}^{2}

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

Step 3.4.2

Simplify each side of the equation.

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Step 3.4.2.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{e^{y} + 1}

$\sqrt{e^{y} + 1}$ as

{(e^{y} + 1)}^{\frac{1}{2}}

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

{({(e^{y} + 1)}^{\frac{1}{2}})}^{2} = {arcsin (x)}^{2}

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

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Simplify

{({(e^{y} + 1)}^{\frac{1}{2}})}^{2}

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

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Step 3.4.2.2.1.1

Multiply the exponents in

{({(e^{y} + 1)}^{\frac{1}{2}})}^{2}

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

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Step 3.4.2.2.1.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

{(e^{y} + 1)}^{\frac{1}{2} \cdot 2} = {arcsin (x)}^{2}

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

Step 3.4.2.2.1.1.2

Cancel the common factor of

2

$2$ .

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Step 3.4.2.2.1.1.2.1

Cancel the common factor.

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

Step 3.4.2.2.1.1.2.2

Rewrite the expression.

${(e^{y} + 1)}^{1} = {\arcsin (x)}^{2}$

Step 3.4.2.2.1.2

Simplify.

$e^{y} + 1 = {\arcsin (x)}^{2}$

Step 3.4.3

Solve for

$y$ .

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Step 3.4.3.1

Subtract

$1$ from both sides of the equation.

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

Step 3.4.3.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{y}) = \ln ({\arcsin (x)}^{2} - 1)$

Step 3.4.3.3

Expand the left side.

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Step 3.4.3.3.1

Expand

$\ln (e^{y})$ by moving

$y$ outside the logarithm.

$y \ln (e) = \ln ({\arcsin (x)}^{2} - 1)$

Step 3.4.3.3.2

The natural logarithm of

$e$ is

$1$ .

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

Step 3.4.3.3.3

Multiply

$y$ by

$1$ .

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

Step 4

Replace

$y$ with

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

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

Step 5

Verify if

$f^{- 1} (x) = \ln ({\arcsin (x)}^{2} - 1)$ is the inverse of

$f (x) = \sin (\sqrt{e^{x} + 1})$ .

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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

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

$f$ into

$f^{- 1}$ .

$f^{- 1} (\sin (\sqrt{e^{x} + 1})) = \ln ({\arcsin (\sin (\sqrt{e^{x} + 1}))}^{2} - 1)$

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

$f (\ln ({\arcsin (x)}^{2} - 1))$ by substituting in the value of

$f^{- 1}$ into

$f$ .

$f (\ln ({\arcsin (x)}^{2} - 1)) = \sin (\sqrt{e^{\ln ({\arcsin (x)}^{2} - 1)} + 1})$

Step 5.3.3

Exponentiation and log are inverse functions.

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

Step 5.3.4

Add

$- 1$ and

$1$ .

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

Step 5.3.5

Add

${\arcsin (x)}^{2}$ and

$0$ .

$f (\ln ({\arcsin (x)}^{2} - 1)) = \sin (\sqrt{{\arcsin (x)}^{2}})$

Step 5.3.6

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

$f (\ln ({\arcsin (x)}^{2} - 1)) = \sin (\arcsin (x))$

Step 5.3.7

The functions sine and arcsine are inverses.

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

Step 5.4

Since

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

$f (f^{- 1} (x)) = x$ , then

$f^{- 1} (x) = \ln ({\arcsin (x)}^{2} - 1)$ is the inverse of

$f (x) = \sin (\sqrt{e^{x} + 1})$ .

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