Calculus Examples

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

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.

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

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify.

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

Step 2.2.2

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

$f (1) = e \cdot 0$

Step 2.2.3

Multiply $e$ by $0$ .

$f (1) = 0$

Step 2.2.4

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.

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

Step 3.2

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

$e^{2} \ln (2)$

Step 3.3

Convert $e^{2} \ln (2)$ to decimal.

$y = 5.1217034$

Step 4

Find the point at $x = 3$ .

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Step 4.1

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

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

Step 4.2

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

$e^{3} \ln (3)$

Step 4.3

Convert $e^{3} \ln (3)$ to decimal.

$y = 22.06621768$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0), (2, 5.1217034), (3, 22.06621768)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0 \\ 2 & 5.122 \\ 3 & 22.066 \end{array}$

Step 6