Precalculus Examples

f (x) = - ln (x - 1) + 3

$f (x) = - \ln (x - 1) + 3$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression

- ln (x - 1) + 3

$- \ln (x - 1) + 3$ is undefined.

x \leq 1

$x \leq 1$

Step 1.2

Since

- ln (x - 1) + 3

$- \ln (x - 1) + 3$

\to

$\to$

\infty

$\infty$ as

x

$x$

\to

$\to$

1

$1$ from the left and

- ln (x - 1) + 3

$- \ln (x - 1) + 3$

\to

$\to$

\infty

$\infty$ as

x

$x$

\to

$\to$

1

$1$ from the right, then

x = 1

$x = 1$ is a vertical asymptote.

x = 1

$x = 1$

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 = 1

$x = 1$

No Horizontal Asymptotes

Vertical Asymptotes:

x = 1

$x = 1$

No Horizontal Asymptotes

Step 2

Find the point at

x = 2

$x = 2$ .

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Step 2.1

Replace the variable

x

$x$ with

2

$2$ in the expression.

f (2) = - ln ((2) - 1) + 3

$f (2) = - \ln ((2) - 1) + 3$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Subtract

1

$1$ from

2

$2$ .

f (2) = - ln (1) + 3

$f (2) = - \ln (1) + 3$

Step 2.2.1.2

The natural logarithm of

1

$1$ is

0

$0$ .

f (2) = - 0 + 3

$f (2) = - 0 + 3$

Step 2.2.1.3

Multiply

- 1

$- 1$ by

0

$0$ .

f (2) = 0 + 3

$f (2) = 0 + 3$

f (2) = 0 + 3

$f (2) = 0 + 3$

Step 2.2.2

Add

0

$0$ and

3

$3$ .

f (2) = 3

$f (2) = 3$

Step 2.2.3

The final answer is

3

$3$ .

3

$3$

3

$3$

Step 2.3

Convert

3

$3$ to decimal.

y = 3

$y = 3$

y = 3

$y = 3$

Step 3

Find the point at

x = 3

$x = 3$ .

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Step 3.1

Replace the variable

x

$x$ with

3

$3$ in the expression.

f (3) = - ln ((3) - 1) + 3

$f (3) = - \ln ((3) - 1) + 3$

Step 3.2

Simplify the result.

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Step 3.2.1

Subtract

1

$1$ from

3

$3$ .

f (3) = - ln (2) + 3

$f (3) = - \ln (2) + 3$

Step 3.2.2

The final answer is

- ln (2) + 3

$- \ln (2) + 3$ .

- ln (2) + 3

$- \ln (2) + 3$

- ln (2) + 3

$- \ln (2) + 3$

Step 3.3

Convert

- ln (2) + 3

$- \ln (2) + 3$ to decimal.

y = 2.30685281

$y = 2.30685281$

y = 2.30685281

$y = 2.30685281$

Step 4

Find the point at

x = 4

$x = 4$ .

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Step 4.1

Replace the variable

x

$x$ with

4

$4$ in the expression.

f (4) = - ln ((4) - 1) + 3

$f (4) = - \ln ((4) - 1) + 3$

Step 4.2

Simplify the result.

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Step 4.2.1

Subtract

1

$1$ from

4

$4$ .

f (4) = - ln (3) + 3

$f (4) = - \ln (3) + 3$

Step 4.2.2

The final answer is

$- \ln (3) + 3$ .

$- \ln (3) + 3$

Step 4.3

Convert

$- \ln (3) + 3$ to decimal.

$y = 1.90138771$

Step 5

The log function can be graphed using the vertical asymptote at

$x = 1$ and the points

$(2, 3), (3, 2.30685281), (4, 1.90138771)$ .

Vertical Asymptote:

$x = 1$

$\begin{array}{c} x & y \\ 2 & 3 \\ 3 & 2.307 \\ 4 & 1.901 \end{array}$

Step 6