Calculus Examples

$\ln (\sqrt{x + 1})$

Step 1

Find the domain for $y = \ln (\sqrt{x + 1})$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 1.1

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

$\sqrt{x + 1} > 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

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

${\sqrt{x + 1}}^{2} > 0^{2}$

Step 1.2.2

Simplify each side of the inequality.

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Step 1.2.2.1

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

${({(x + 1)}^{\frac{1}{2}})}^{2} > 0^{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

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

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Step 1.2.2.2.1.1

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

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Step 1.2.2.2.1.1.1

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

${(x + 1)}^{\frac{1}{2} \cdot 2} > 0^{2}$

Step 1.2.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.1.2.1

Cancel the common factor.

${(x + 1)}^{\frac{1}{2} \cdot 2} > 0^{2}$

Step 1.2.2.2.1.1.2.2

Rewrite the expression.

${(x + 1)}^{1} > 0^{2}$

Step 1.2.2.2.1.2

Simplify.

$x + 1 > 0^{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

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

$x + 1 > 0$

Step 1.2.3

Subtract $1$ from both sides of the inequality.

$x > - 1$

Step 1.2.4

Find the domain of $\sqrt{x + 1}$ .

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Step 1.2.4.1

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

$x + 1 \geq 0$

Step 1.2.4.2

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 1.2.4.3

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

$[- 1, \infty)$

Step 1.2.5

The solution consists of all of the true intervals.

$x > - 1$

Step 1.3

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

$x + 1 \geq 0$

Step 1.4

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

Step 1.5

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

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Interval Notation:

$(- 1, \infty)$

Set-Builder Notation:

${x | x > - 1}$

Step 2

To find the radical expression end point, substitute the $x$ value $- 1$ , which is the least value in the domain, into $f (x) = \ln (\sqrt{x + 1})$ .

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Step 2.1

Replace the variable $x$ with $- 1$ in the expression.

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

Step 2.2

Add $- 1$ and $1$ .

$f (- 1) = \ln (\sqrt{0})$

Step 2.3

Rewrite $0$ as $0^{2}$ .

$f (- 1) = \ln (\sqrt{0^{2}})$

Step 2.4

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

$f (- 1) = \ln (0)$

Step 2.5

The natural logarithm of zero is undefined.

$f (- 1) = Undefined$

Undefined

Step 3

The radical expression end point is $(- 1, Undefined)$ .

$(- 1, Undefined)$

Step 4

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

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Step 4.1

Substitute the $x$ value $0$ into $f (x) = \ln (\sqrt{x + 1})$ . In this case, the point is $(0, 0)$ .

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Step 4.1.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \ln (\sqrt{(0) + 1})$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Add $0$ and $1$ .

$f (0) = \ln (\sqrt{1})$

Step 4.1.2.2

Any root of $1$ is $1$ .

$f (0) = \ln (1)$

Step 4.1.2.3

The natural logarithm of $1$ is $0$ .

$f (0) = 0$

Step 4.1.2.4

The final answer is $0$ .

$y = 0$

Step 4.2

Substitute the $x$ value $1$ into $f (x) = \ln (\sqrt{x + 1})$ . In this case, the point is $(1, \ln (\sqrt{2}))$ .

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Step 4.2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln (\sqrt{(1) + 1})$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Add $1$ and $1$ .

$f (1) = \ln (\sqrt{2})$

Step 4.2.2.2

The final answer is $\ln (\sqrt{2})$ .

$y = \ln (\sqrt{2})$

Step 4.3

The square root can be graphed using the points around the vertex $(- 1, Undefined), (0, 0), (1, 0.35)$

$\begin{array}{c} x & y \\ - 1 & Undefined \\ 0 & 0 \\ 1 & 0.35 \end{array}$

Step 5