Precalculus Examples

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

Step 1

Set the denominator in $\frac{1 + x^{2}}{2 - e^{3 x + 1}}$ equal to $0$ to find where the expression is undefined.

$2 - e^{3 x + 1} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $2$ from both sides of the equation.

$- e^{3 x + 1} = - 2$

Step 2.2

Divide each term in $- e^{3 x + 1} = - 2$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- e^{3 x + 1} = - 2$ by $- 1$ .

$\frac{- e^{3 x + 1}}{- 1} = \frac{- 2}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{e^{3 x + 1}}{1} = \frac{- 2}{- 1}$

Step 2.2.2.2

Divide $e^{3 x + 1}$ by $1$ .

$e^{3 x + 1} = \frac{- 2}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 2$ by $- 1$ .

$e^{3 x + 1} = 2$

Step 2.3

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

$\ln (e^{3 x + 1}) = \ln (2)$

Step 2.4

Expand the left side.

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Step 2.4.1

Expand $\ln (e^{3 x + 1})$ by moving $3 x + 1$ outside the logarithm.

$(3 x + 1) \ln (e) = \ln (2)$

Step 2.4.2

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

$(3 x + 1) \cdot 1 = \ln (2)$

Step 2.4.3

Multiply $3 x + 1$ by $1$ .

$3 x + 1 = \ln (2)$

Step 2.5

Subtract $1$ from both sides of the equation.

$3 x = \ln (2) - 1$

Step 2.6

Divide each term in $3 x = \ln (2) - 1$ by $3$ and simplify.

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Step 2.6.1

Divide each term in $3 x = \ln (2) - 1$ by $3$ .

$\frac{3 x}{3} = \frac{\ln (2)}{3} + \frac{- 1}{3}$

Step 2.6.2

Simplify the left side.

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Step 2.6.2.1

Cancel the common factor of $3$ .

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Step 2.6.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\ln (2)}{3} + \frac{- 1}{3}$

Step 2.6.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (2)}{3} + \frac{- 1}{3}$

Step 2.6.3

Simplify the right side.

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Step 2.6.3.1

Move the negative in front of the fraction.

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

Step 3

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

Interval Notation:

$(- \infty, \frac{\ln (2)}{3} - \frac{1}{3}) \cup (\frac{\ln (2)}{3} - \frac{1}{3}, \infty)$

Set-Builder Notation:

${x | x \neq \frac{\ln (2)}{3} - \frac{1}{3}}$

Step 4