Calculus Examples

$\ln (x e^{\sqrt{x}} + 2)$

Step 1

Find the domain for $y = \ln (x e^{\sqrt{x}} + 2)$ 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 $\ln (x e^{\sqrt{x}} + 2)$ greater than $0$ to find where the expression is defined.

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

Step 1.2

Solve for $x$ .

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Step 1.2.1

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

Step 1.2.2

Find the domain of $x e^{\sqrt{x}} + 2$ .

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Step 1.2.2.1

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

$x \geq 0$

Step 1.2.2.2

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

$[0, \infty)$

Step 1.2.3

The solution consists of all of the true intervals.

$x > 0$

Step 1.3

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

$x \geq 0$

Step 1.4

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

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 2

To find the radical expression end point, substitute the $x$ value $0$ , which is the least value in the domain, into $f (x) = \ln (x e^{\sqrt{x}} + 2)$ .

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Step 2.1

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

$f (0) = \ln ((0) \cdot e^{\sqrt{0}} + 2)$

Step 2.2

Simplify the result.

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Step 2.2.1

Remove parentheses.

$f (0) = \ln ((0) \cdot e^{\sqrt{0}} + 2)$

Step 2.2.2

Simplify each term.

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Step 2.2.2.1

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

$f (0) = \ln (0 \cdot e^{\sqrt{0^{2}}} + 2)$

Step 2.2.2.2

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

$f (0) = \ln (0 \cdot e^{0} + 2)$

Step 2.2.2.3

Anything raised to $0$ is $1$ .

$f (0) = \ln (0 \cdot 1 + 2)$

Step 2.2.2.4

Multiply $0$ by $1$ .

$f (0) = \ln (0 + 2)$

Step 2.2.3

Add $0$ and $2$ .

$f (0) = \ln (2)$

Step 2.2.4

The final answer is $\ln (2)$ .

$\ln (2)$

Step 3

The radical expression end point is $(0, \ln (2))$ .

$(0, \ln (2))$

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

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Step 4.1.1

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

$f (1) = \ln ((1) \cdot e^{\sqrt{1}} + 2)$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Remove parentheses.

$f (1) = \ln ((1) \cdot e^{\sqrt{1}} + 2)$

Step 4.1.2.2

Simplify each term.

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Step 4.1.2.2.1

Multiply $e^{\sqrt{1}}$ by $1$ .

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

Step 4.1.2.2.2

Any root of $1$ is $1$ .

$f (1) = \ln (e + 2)$

Step 4.1.2.2.3

Simplify.

$f (1) = \ln (e + 2)$

Step 4.1.2.3

The final answer is $\ln (e + 2)$ .

$y = \ln (e + 2)$

Step 4.2

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

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Step 4.2.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \ln ((2) \cdot e^{\sqrt{2}} + 2)$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Remove parentheses.

$f (2) = \ln ((2) \cdot e^{\sqrt{2}} + 2)$

Step 4.2.2.2

Multiply $2$ by $e^{\sqrt{2}}$ .

$f (2) = \ln (2 e^{\sqrt{2}} + 2)$

Step 4.2.2.3

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

$y = \ln (2 e^{\sqrt{2}} + 2)$

Step 4.3

The square root can be graphed using the points around the vertex $(0, \ln (2)), (1, 1.55), (2, 2.32)$

$\begin{array}{c} x & y \\ 0 & \ln (2) \\ 1 & 1.55 \\ 2 & 2.32 \end{array}$

Step 5