Calculus Examples

$g (x) = \sqrt{\ln (x - 1)}$

Step 1

Set the argument in $\ln (x - 1)$ greater than $0$ to find where the expression is defined.

$x - 1 > 0$

Step 2

Add $1$ to both sides of the inequality.

$x > 1$

Step 3

Set the radicand in $\sqrt{\ln (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$\ln (x - 1) \geq 0$

Step 4

Solve for $x$ .

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Step 4.1

Convert the inequality to an equality.

$\ln (x - 1) = 0$

Step 4.2

Solve the equation.

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Step 4.2.1

To solve for $x$ , rewrite the equation using properties of logarithms.

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

Step 4.2.2

Rewrite $\ln (x - 1) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = x - 1$

Step 4.2.3

Solve for $x$ .

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Step 4.2.3.1

Rewrite the equation as $x - 1 = e^{0}$ .

$x - 1 = e^{0}$

Step 4.2.3.2

Anything raised to $0$ is $1$ .

$x - 1 = 1$

Step 4.2.3.3

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

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Step 4.2.3.3.1

Add $1$ to both sides of the equation.

$x = 1 + 1$

Step 4.2.3.3.2

Add $1$ and $1$ .

$x = 2$

Step 4.3

Find the domain of $\ln (x - 1)$ .

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Step 4.3.1

Set the argument in $\ln (x - 1)$ greater than $0$ to find where the expression is defined.

$x - 1 > 0$

Step 4.3.2

Add $1$ to both sides of the inequality.

$x > 1$

Step 4.3.3

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

$(1, \infty)$

Step 4.4

The solution consists of all of the true intervals.

$x \geq 2$

Step 5

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

Interval Notation:

$[2, \infty)$

Set-Builder Notation:

${x | x \geq 2}$

Step 6

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${y | y \geq 0}$

Step 7

Determine the domain and range.

Domain: $[2, \infty), {x | x \geq 2}$

Range: $[0, \infty), {y | y \geq 0}$

Step 8