Graph each function. ƒ(x) = log3 (x-1) + 2

Ch. 4 - Inverse, Exponential, and Logarithmic Functions

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Lial 13th Edition

Ch. 4 - Inverse, Exponential, and Logarithmic Functions

문제 55

5장, 문제 55

Graph each function. ƒ(x) = log₃ (x-1) + 2

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Identify the base of the logarithm and the transformations applied to the function. Here, the function is \(f(x) = \log_{3}(x - 1) + 2\), which is a logarithmic function with base 3, shifted horizontally and vertically.

Determine the domain of the function by setting the argument of the logarithm greater than zero: \(x - 1 > 0\). Solve this inequality to find the domain.

Find the vertical asymptote by setting the argument of the logarithm equal to zero: \(x - 1 = 0\). This vertical line is where the function approaches infinity or negative infinity.

Identify the horizontal shift and vertical shift. The \(x - 1\) inside the logarithm shifts the graph 1 unit to the right, and the \(+2\) outside shifts the graph 2 units upward.

Plot key points by choosing values of \(x\) greater than 1, calculate corresponding \(f(x)\) values using the function, and sketch the graph showing the vertical asymptote at \(x=1\), the general shape of the logarithmic curve, and the shifts.

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Logarithmic Functions

A logarithmic function is the inverse of an exponential function and is written as f(x) = log_b(x), where b is the base. It answers the question: to what power must the base b be raised to produce x? Understanding the properties of logarithms is essential for graphing and interpreting these functions.

Domain of Logarithmic Functions

The domain of a logarithmic function f(x) = log_b(x - h) is all x-values for which the argument (x - h) is positive. For example, in f(x) = log_3(x - 1) + 2, the domain is x > 1 because the logarithm is undefined for zero or negative inputs. Identifying the domain helps in correctly plotting the graph.

Transformations of Functions

Transformations shift or change the shape of a graph. For f(x) = log_3(x - 1) + 2, the (x - 1) inside the log shifts the graph 1 unit to the right, and the +2 outside shifts it 2 units up. Recognizing horizontal and vertical shifts is crucial for accurately graphing the function.