Problem 77 - Precalculus Mathematics for Calculus

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

\(f \circ g(x) = \log_2(x-2)\) on \((2, \infty)\); \(g \circ f(x) = \log_2(x) - 2\) on \((0, \infty)\).

Step 1: Understand the Problem

We need to find the composite functions \(f \circ g\) and \(g \circ f\) using the given functions \(f(x) = \log_2{x}\) and \(g(x) = x - 2\). We also need to determine the domains of these composite functions.

Step 2: Find \(f \circ g\)

The notation \(f \circ g\) means the function \(f\) composed with \(g\), which means we plug \(g(x)\) into \(f(x)\). So, \((f \circ g)(x) = f(g(x)) = f(x - 2) = \log_2{(x-2)}\).

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

For \(f(x) = \log_2{x}\), the domain is \(x > 0\). Therefore, for \(\log_2{(x-2)}\) to be defined, \(x - 2 > 0\), which simplifies to \(x > 2\). So, the domain of \(f \circ g\) is \((2, \infty)\).

Step 4: Find \(g \circ f\)

The notation \(g \circ f\) means the function \(g\) composed with \(f\), which means we plug \(f(x)\) into \(g(x)\). So, \((g \circ f)(x) = g(f(x)) = g(\log_2{x}) = \log_2{x} - 2\).

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

For \(f(x) = \log_2{x}\), the domain is \(x > 0\) (since \(\log_2{x}\) is defined for positive \(x\)). Since \(g(x) = x - 2\) is defined for all real numbers, the domain of \(g \circ f\) is simply \(x > 0\).

Step 6: Conclusion

The composite functions are \(f \circ g(x) = \log_2(x-2)\) with domain \((2, \infty)\) and \(g \circ f(x) = \log_2(x) - 2\) with domain \((0, \infty)\).

Key Concepts

Domain of a function

The domain of a function is the complete set of all possible input values (usually denoted as \(x\)) for which the function is well-defined. To determine the domain of a function, consider where the function input makes the function meaningful or avoids mathematical errors.

• For example, the function \(f(x) = \log_2{x}\) is only defined for positive \(x\), so its domain is \((0, \,\infty)\).
• Adding 2 to \(x\) and then computing \(\log_2{(x-2)}\) means the input \(x-2\) must be greater than zero. So, \(x > 2\).

To find the domain of composite functions, analyze each part separately and consider combined constraints. Always start with the inside function, and build the domain step by step, ensuring each part is correctly defined.
This method helps to understand limits and boundaries of function inputs accurately.

Logarithmic functions

Logarithmic functions map numbers to their corresponding exponent in a given base. The function \(f(x) = \log_b{x}\) essentially answers "What power \(b\) must be raised to, to obtain \(x\)?" For \(\log_2{x}\), it results in the power 2 must be raised to equal \(x\).

• Logarithmic functions invert exponential functions. For instance, if \(2^3 = 8\), then \(\log_2{8} = 3\).
• The base of logarithms (in this case 2) determines the scaling. Different bases shift how the function grows or shrinks.
• The natural logarithm (written as \(\ln\)) is a common variant, using base \(e\), an irrational constant near 2.718.

The domain of log functions is always limited to positive values of \(x\). Negative numbers and zero do not have real-number logarithms, because no real number power of a positive base produces negative or zero outcomes.

Function composition

Function composition involves combining two functions, such that the output of one function becomes the input of another. When you see \(f \circ g\), read it as "\(f\) composed with \(g\)". It means \(f(g(x))\), plugging \(g(x)\) into \(f\).

• For \(f(x) = \log_2{x}\) and \(g(x) = x - 2\), \(f \circ g(x) = \log_2(x - 2)\) requires \(x - 2 > 0\), giving a domain of \((2, \,\infty)\).
• Reversing to \(g \circ f = g(f(x))\) leads to \(g \circ f(x) = \log_2x - 2\) with input restrictions \(\log_2x\) needing \(x > 0\), offering a domain of \((0, \,\infty)\).
• Function composition is highly useful, allowing complex operations through simpler building blocks.

When working with composites, always align compatible domains to ensure the entire function is valid. It synthesizes distinct concepts into a unified framework, opening new possibilities in problem-solving.