Calculus Examples

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

Step 1

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

$x > 0$

Step 1.2

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

$x \geq 0$

Step 1.3

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 $f (x) = \sqrt{x} \ln (x)$ .

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Step 2.1

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

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

Step 2.2

Remove parentheses.

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

Step 2.3

The natural logarithm of zero is undefined.

$f (0) = Undefined$

Undefined

Step 3

The radical expression end point is $(0, Undefined)$ .

$(0, Undefined)$

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Remove parentheses.

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

Step 4.1.2.2

Any root of $1$ is $1$ .

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

Step 4.1.2.3

Multiply $\ln (1)$ by $1$ .

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

Step 4.1.2.4

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

$f (1) = 0$

Step 4.1.2.5

The final answer is $0$ .

$y = 0$

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Remove parentheses.

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

Step 4.2.2.2

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

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

Step 4.3

The square root can be graphed using the points around the vertex $(0, Undefined), (1, 0), (2, 0.98)$

$\begin{array}{c} x & y \\ 0 & Undefined \\ 1 & 0 \\ 2 & 0.98 \end{array}$

Step 5