Calculus Examples

\ln (x e^{x})

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

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression

\ln (x) + x

$\ln (x) + x$ is undefined.

x \leq 0

$x \leq 0$

Step 1.2

Since

\ln (x) + x

$\ln (x) + x$

\to

$\to$

\infty

$\infty$ as

x

$x$

\to

$\to$

0

$0$ from the left and

\ln (x) + x

$\ln (x) + x$

\to

$\to$

- \infty

$- \infty$ as

x

$x$

\to

$\to$

0

$0$ from the right, then

x = 0

$x = 0$ is a vertical asymptote.

x = 0

$x = 0$

Step 1.3

Ignoring the logarithm, consider the rational function

R (x) = \frac{a x^{n}}{b x^{m}}

$R (x) = \frac{a x^{n}}{b x^{m}}$ where

n

$n$ is the degree of the numerator and

m

$m$ is the degree of the denominator.

1. If

n < m

$n < m$ , then the x-axis,

y = 0

$y = 0$ , is the horizontal asymptote.

2. If

n = m

$n = m$ , then the horizontal asymptote is the line

y = \frac{a}{b}

$y = \frac{a}{b}$ .

3. If

n > m

$n > m$ , then there is no horizontal asymptote (there is an oblique asymptote).

Step 1.4

There are no horizontal asymptotes because

Q (x)

$Q (x)$ is

1

$1$ .

No Horizontal Asymptotes

Step 1.5

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.6

This is the set of all asymptotes.

Vertical Asymptotes:

x = 0

$x = 0$

No Horizontal Asymptotes

Vertical Asymptotes:

x = 0

$x = 0$

No Horizontal Asymptotes

Step 2

Find the point at

x = 1

$x = 1$ .

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Step 2.1

Replace the variable

x

$x$ with

1

$1$ in the expression.

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

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

Step 2.2

Simplify the result.

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Step 2.2.1

Multiply

e^{1}

$e^{1}$ by

1

$1$ .

f (1) = \ln (e)

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

Step 2.2.2

Use logarithm rules to move

1

$1$ out of the exponent.

f (1) = 1 \ln (e)

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

Step 2.2.3

Multiply

\ln (e)

$\ln (e)$ by

1

$1$ .

f (1) = \ln (e)

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

Step 2.2.4

The natural logarithm of

e

$e$ is

1

$1$ .

f (1) = 1

$f (1) = 1$

Step 2.2.5

The final answer is

1

$1$ .

1

$1$

1

$1$

Step 2.3

Convert

1

$1$ to decimal.

y = 1

$y = 1$

y = 1

$y = 1$

Step 3

Find the point at

x = 2

$x = 2$ .

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Step 3.1

Replace the variable

x

$x$ with

2

$2$ in the expression.

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Multiply

2

$2$ by

e^{2}

$e^{2}$ .

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

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

Step 3.2.2

The final answer is

\ln (2 e^{2})

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

\ln (2 e^{2})

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

\ln (2 e^{2})

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

Step 3.3

Convert

\ln (2 e^{2})

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

y = 2.69314718

$y = 2.69314718$

y = 2.69314718

$y = 2.69314718$

Step 4

Find the point at

x = 3

$x = 3$ .

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Step 4.1

Replace the variable

x

$x$ with

3

$3$ in the expression.

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

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

Step 4.2

Simplify the result.

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Step 4.2.1

Multiply

3

$3$ by

e^{3}

$e^{3}$ .

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

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

Step 4.2.2

The final answer is

\ln (3 e^{3})

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

\ln (3 e^{3})

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

\ln (3 e^{3})

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

Step 4.3

Convert

\ln (3 e^{3})

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

y = 4.09861228

$y = 4.09861228$

y = 4.09861228

$y = 4.09861228$

Step 5

The log function can be graphed using the vertical asymptote at

x = 0

$x = 0$ and the points

(1, 1), (2, 2.69314718), (3, 4.09861228)

$(1, 1), (2, 2.69314718), (3, 4.09861228)$ .

Vertical Asymptote:

x = 0

$x = 0$

\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 2.693 \\ 3 & 4.099 \end{array}

$\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 2.693 \\ 3 & 4.099 \end{array}$

Step 6