Question

use the given logarithmic function to answer the following questions.
$f(x) = \ln(x - 1) - 3$

a) what are the domain and range of $f$?

domain

range

b) what is the $x$-intercept of the graph of $f$? enter your answer as an ordered pair. if no $x$-intercept exists, enter none.

$x$-intercept

c) the graph of $\ln(x)$ is shown below. sketch the graph of $f(x) = \ln(x - 1) - 3$ by entering values for the horizontal/vertical shift(s) in the boxes below, then clicking \graph\.

Explanation:

Part (a)

Step1: Find Domain of \( f(x) = \ln(x - 1) - 3 \)

The argument of a natural logarithm \( \ln(u) \) must be positive, so \( x - 1 > 0 \). Solving this inequality: \( x > 1 \). Thus, the domain is all real numbers greater than 1.

Step2: Find Range of \( f(x) = \ln(x - 1) - 3 \)

The range of \( \ln(u) \) is \( (-\infty, \infty) \). Shifting the graph vertically by -3 (subtracting 3) does not change the range. So the range is all real numbers, \( (-\infty, \infty) \).

Part (b)

Step1: Find \( x \)-intercept (set \( f(x) = 0 \))

Set \( \ln(x - 1) - 3 = 0 \). Add 3 to both sides: \( \ln(x - 1) = 3 \).

Step2: Solve for \( x \)

Exponentiate both sides with base \( e \): \( e^{\ln(x - 1)} = e^3 \). Simplify: \( x - 1 = e^3 \). Add 1: \( x = e^3 + 1 \). The \( x \)-intercept is \( (e^3 + 1, 0) \) (since \( y = 0 \) at the \( x \)-intercept).

Part (c)

Step1: Horizontal Shift

For \( f(x) = \ln(x - h) + k \), the horizontal shift is \( h \). Here, \( f(x) = \ln(x - 1) - 3 \), so \( h = 1 \) (shift 1 unit to the right).

Step2: Vertical Shift

The vertical shift is \( k \). Here, \( k = -3 \) (shift 3 units down).

Part (a)

Answer:

Domain: \( (1, \infty) \)
Range: \( (-\infty, \infty) \)

Part (b)