Trigonometry Examples

$x \ln (x)$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$x = 0$

Step 1.2

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

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

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

Step 2.2.2

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

$f (1) = 0$

Step 2.2.3

The final answer is $0$ .

$0$

Step 2.3

Convert $0$ to decimal.

$y = 0$

Step 3

Find the point at $x = 2$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify $(2) \ln (2)$ by moving $2$ inside the logarithm.

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

Step 3.2.2

Raise $2$ to the power of $2$ .

$f (2) = \ln (4)$

Step 3.2.3

The final answer is $\ln (4)$ .

$\ln (4)$

Step 3.3

Convert $\ln (4)$ to decimal.

$y = 1.38629436$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

$f (3) = (3) \ln (3)$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify $(3) \ln (3)$ by moving $3$ inside the logarithm.

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

Step 4.2.2

Raise $3$ to the power of $3$ .

$f (3) = \ln (27)$

Step 4.2.3

The final answer is $\ln (27)$ .

$\ln (27)$

Step 4.3

Convert $\ln (27)$ to decimal.

$y = 3.29583686$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (2, 1.38629436), (3, 3.29583686)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 1.386 \\ 3 & 3.296 \end{array}$

Step 6