Calculus Examples

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

Step 1

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

$\ln (x) \geq 0$

Step 1.3

Solve for $x$ .

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Step 1.3.1

Convert the inequality to an equality.

$\ln (x) = 0$

Step 1.3.2

Solve the equation.

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Step 1.3.2.1

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

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

Step 1.3.2.2

Rewrite $\ln (x) = 0$ 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 $b^{y} = x$ .

$e^{0} = x$

Step 1.3.2.3

Solve for $x$ .

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Step 1.3.2.3.1

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

$x = e^{0}$

Step 1.3.2.3.2

Anything raised to $0$ is $1$ .

$x = 1$

Step 1.3.3

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

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Step 1.3.3.1

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

$x > 0$

Step 1.3.3.2

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

$(0, \infty)$

Step 1.3.4

The solution consists of all of the true intervals.

$x \geq 1$

Step 1.4

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

Interval Notation:

$[1, \infty)$

Set-Builder Notation:

${x | x \geq 1}$

Interval Notation:

$[1, \infty)$

Set-Builder Notation:

${x | x \geq 1}$

Step 2

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

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Multiply $\sqrt{\ln (1)}$ by $1$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

$f (1) = 0$

Step 2.2.5

The final answer is $0$ .

$0$

Step 3

The radical expression end point is $(1, 0)$ .

$(1, 0)$

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Multiply $2$ by $\sqrt{\ln (2)}$ .

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

Step 4.1.2.2

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

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

Step 4.2

Substitute the $x$ value $3$ into $f (x) = x \sqrt{\ln (x)}$ . In this case, the point is $(3, 3 \sqrt{\ln (3)})$ .

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Step 4.2.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = (3) \sqrt{\ln (3)}$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Multiply $3$ by $\sqrt{\ln (3)}$ .

$f (3) = 3 \sqrt{\ln (3)}$

Step 4.2.2.2

The final answer is $3 \sqrt{\ln (3)}$ .

$y = 3 \sqrt{\ln (3)}$

Step 4.3

The square root can be graphed using the points around the vertex $(1, 0), (2, 1.67), (3, 3.14)$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 1.67 \\ 3 & 3.14 \end{array}$

Step 5