Precalculus Examples

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

Step 1

Set the radicand in $\sqrt{e^{2 x} - e^{x} - 2}$ greater than or equal to $0$ to find where the expression is defined.

$e^{2 x} - e^{x} - 2 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Rewrite $e^{2 x}$ as exponentiation.

${(e^{x})}^{2} - e^{x} - 2 = 0$

Step 2.2

Substitute $u$ for $e^{x}$ .

$u^{2} - u - 2 = 0$

Step 2.3

Solve for $u$ .

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Step 2.3.1

Factor $u^{2} - u - 2$ using the AC method.

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Step 2.3.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 2.3.1.2

Write the factored form using these integers.

$(u - 2) (u + 1) = 0$

Step 2.3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u - 2 = 0$

$u + 1 = 0$

Step 2.3.3

Set $u - 2$ equal to $0$ and solve for $u$ .

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Step 2.3.3.1

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Step 2.3.3.2

Add $2$ to both sides of the equation.

$u = 2$

Step 2.3.4

Set $u + 1$ equal to $0$ and solve for $u$ .

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Step 2.3.4.1

Set $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 2.3.4.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 2.3.5

The final solution is all the values that make $(u - 2) (u + 1) = 0$ true.

$u = 2, - 1$

Step 2.4

Substitute $2$ for $u$ in $u = e^{x}$ .

$2 = e^{x}$

Step 2.5

Solve $2 = e^{x}$ .

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Step 2.5.1

Rewrite the equation as $e^{x} = 2$ .

$e^{x} = 2$

Step 2.5.2

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

$\ln (e^{x}) = \ln (2)$

Step 2.5.3

Expand the left side.

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Step 2.5.3.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (2)$

Step 2.5.3.2

The natural logarithm of $e$ is $1$ .

$x \cdot 1 = \ln (2)$

Step 2.5.3.3

Multiply $x$ by $1$ .

$x = \ln (2)$

Step 2.6

Substitute $- 1$ for $u$ in $u = e^{x}$ .

$- 1 = e^{x}$

Step 2.7

Solve $- 1 = e^{x}$ .

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Step 2.7.1

Rewrite the equation as $e^{x} = - 1$ .

$e^{x} = - 1$

Step 2.7.2

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

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

Step 2.7.3

The equation cannot be solved because $\ln (- 1)$ is undefined.

Undefined

Step 2.7.4

There is no solution for $e^{x} = - 1$

No solution

Step 2.8

List the solutions that makes the equation true.

$x = \ln (2)$

Step 2.9

The solution consists of all of the true intervals.

$x \geq \ln (2)$

Step 3

Set the denominator in $\frac{e^{- x}}{\sqrt{e^{2 x} - e^{x} - 2}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{e^{2 x} - e^{x} - 2} = 0$

Step 4

Solve for $x$ .

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Step 4.1

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

${\sqrt{e^{2 x} - e^{x} - 2}}^{2} = 0^{2}$

Step 4.2

Simplify each side of the equation.

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Step 4.2.1

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

${({(e^{2 x} - e^{x} - 2)}^{\frac{1}{2}})}^{2} = 0^{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

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

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Step 4.2.2.1.1

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

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Step 4.2.2.1.1.1

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

${(e^{2 x} - e^{x} - 2)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.2.2.1.1.2.1

Cancel the common factor.

${(e^{2 x} - e^{x} - 2)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 4.2.2.1.1.2.2

Rewrite the expression.

${(e^{2 x} - e^{x} - 2)}^{1} = 0^{2}$

Step 4.2.2.1.2

Simplify.

$e^{2 x} - e^{x} - 2 = 0^{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Raising $0$ to any positive power yields $0$ .

$e^{2 x} - e^{x} - 2 = 0$

Step 4.3

Solve for $x$ .

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Step 4.3.1

Factor the left side of the equation.

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Step 4.3.1.1

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

${(e^{x})}^{2} - e^{x} - 2 = 0$

Step 4.3.1.2

Let $u = e^{x}$ . Substitute $u$ for all occurrences of $e^{x}$ .

$u^{2} - u - 2 = 0$

Step 4.3.1.3

Factor $u^{2} - u - 2$ using the AC method.

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Step 4.3.1.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 4.3.1.3.2

Write the factored form using these integers.

$(u - 2) (u + 1) = 0$

Step 4.3.1.4

Replace all occurrences of $u$ with $e^{x}$ .

$(e^{x} - 2) (e^{x} + 1) = 0$

Step 4.3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{x} - 2 = 0$

$e^{x} + 1 = 0$

Step 4.3.3

Set $e^{x} - 2$ equal to $0$ and solve for $x$ .

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Step 4.3.3.1

Set $e^{x} - 2$ equal to $0$ .

$e^{x} - 2 = 0$

Step 4.3.3.2

Solve $e^{x} - 2 = 0$ for $x$ .

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Step 4.3.3.2.1

Add $2$ to both sides of the equation.

$e^{x} = 2$

Step 4.3.3.2.2

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

$\ln (e^{x}) = \ln (2)$

Step 4.3.3.2.3

Expand the left side.

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Step 4.3.3.2.3.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (2)$

Step 4.3.3.2.3.2

The natural logarithm of $e$ is $1$ .

$x \cdot 1 = \ln (2)$

Step 4.3.3.2.3.3

Multiply $x$ by $1$ .

$x = \ln (2)$

Step 4.3.4

Set $e^{x} + 1$ equal to $0$ and solve for $x$ .

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Step 4.3.4.1

Set $e^{x} + 1$ equal to $0$ .

$e^{x} + 1 = 0$

Step 4.3.4.2

Solve $e^{x} + 1 = 0$ for $x$ .

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Step 4.3.4.2.1

Subtract $1$ from both sides of the equation.

$e^{x} = - 1$

Step 4.3.4.2.2

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

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

Step 4.3.4.2.3

The equation cannot be solved because $\ln (- 1)$ is undefined.

Undefined

Step 4.3.4.2.4

There is no solution for $e^{x} = - 1$

No solution

Step 4.3.5

The final solution is all the values that make $(e^{x} - 2) (e^{x} + 1) = 0$ true.

$x = \ln (2)$

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(\ln (2), \infty)$

Set-Builder Notation:

${x | x > \ln (2)}$

Step 6