Problem 77 - Algebra and Trigonometry

Question

Find the functions \(f \circ g\) and \(g \circ f\) and their domains. $$ f(x)=\log _{2} x, \quad g(x)=x-2 $$

Step-by-Step Solution

Answer

\(f \circ g(x) = \log_2 (x-2)\), domain \(x > 2\); \(g \circ f(x) = \log_2 x - 2\), domain \(x > 0\).

Step 1: Understanding Function Composition

Function composition is when you apply one function to the result of another function. For example, for the function composition \((f \circ g)(x)\), you apply \(g(x)\) first, and then apply \(f(x)\) to the result. Conversely, for \((g \circ f)(x)\), apply \(f(x)\) first, then \(g(x)\) to the result.

Step 2: Finding \(f \circ g(x)\)

To find \((f \circ g)(x)\), substitute \(g(x) = x-2\) into \(f(x) = \log_2 x\). So, \(f(g(x)) = f(x-2) = \log_2 (x-2)\).

Step 3: Finding the Domain of \(f \circ g(x)\)

The domain of \(f(x) = \log_2 x\) is \(x > 0\). For \(f \circ g(x) = \log_2 (x-2)\), we need \(x - 2 > 0\). Solving this inequality, we get \(x > 2\). Thus, the domain of \(f \circ g(x)\) is \(x > 2\).

Step 4: Finding \(g \circ f(x)\)

To find \((g \circ f)(x)\), substitute \(f(x) = \log_2 x\) into \(g(x) = x - 2\). So, \(g(f(x)) = g(\log_2 x) = \log_2 x - 2\).

Step 5: Finding the Domain of \(g \circ f(x)\)

The domain of \(g \circ f(x) = \log_2 x - 2\) is determined by the domain of \(f(x) = \log_2 x\), which requires \(x > 0\). Therefore, the domain of \(g \circ f(x)\) is \(x > 0\).

Key Concepts

Understanding Domains of Functions

The domain of a function is all the possible values of 'x' that you can input into the function without causing any mathematical issue. For a simple function like a linear one, such as \(g(x) = x-2\), the domain can be any real number. This is because you can subtract 2 from any number without entering undefined territory.
However, with functions like \(f(x) = \log_2 x\), the domain is restricted to \(x > 0\). That's because the logarithmic function, \(\log_2 x\), is only defined for positive numbers. When dealing with the composition \(f \circ g(x)\), which is \(\log_2 (x-2)\), you need to make sure that the expression inside the logarithm, \(x-2\), is greater than zero. Solving that inequality gives you \(x > 2\), establishing the domain for this composition.
Similarly, for \(g \circ f(x) = \log_2 x - 2\), the domain is directly influenced by the first function, \(f(x) = \log_2 x\), meaning \(x > 0\). Understanding domains helps ensure the calculations are valid and the functions behave as expected.

Exploring Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They answer the question, "to what power must the base be raised, to produce a given number?" Here, we deal with \(f(x) = \log_2 x\), which can be understood as 'what power should 2 be to get x?'
For logarithmic functions:

• The function is only defined for positive numbers.
• The base must be a positive number greater than 1, here it's 2, but logarithms can have any base.
• The graph of a logarithmic function steadily rises, crossing the x-axis at (1,0).

When composing a logarithmic function as in \(f \circ g(x) = \log_2 (x-2)\), ensure the input remains positive by solving \(x-2 > 0\). Logarithmic rules maintain predictable behavior within their valid domains.

Mastering Inequality Solving

Solving inequalities is crucial for determining domains, especially when working with functions that have restrictions, like logarithms. Inequalities are statements about the relative size or order of two objects or numbers.
For function domains involving inequalities:

• To find where \(f(x) = \log_2 (x-2)\) is defined, solve \(x-2 > 0\). Easy, isn't it?
• Simply add 2 to both sides to isolate x, resulting in \(x > 2\).

This step ensures that the value inside the logarithm is positive, making the function valid. Remember, when solving inequalities, flip the inequality sign whenever you multiply or divide by a negative number. However, this exercise doesn't require that.
Mastering these steps helps enhance problem-solving skills across various math topics.