Question

If \(f(x)=\log _{2} x, g(x)=2^{x}\) and \(h(x)=4 x,\) find: (a) \((f \circ g)(x) .\) What is the domain of \(f \circ g ?\) (b) \((g \circ f)(x) .\) What is the domain of \(g \circ f ?\) (c) \((f \circ g)(3)\) (d) \((f \circ h)(x) .\) What is the domain of \(f \circ h ?\) (e) \((f \circ h)(8)\)

Short answer

(a) \((f \circ g)(x) = x\); domain: \(( -\infty, \infty)\). (b) \((g \circ f)(x) = x\); domain: \((0, \infty)\). (c) \((f \circ g)(3) = 3\). (d) \((f \circ h)(x) = \log_{2}(4x)\); domain: \((0, \infty)\). (e) \((f \circ h)(8) = 5\).

Step-by-step solution

Understanding Function Composition

Function composition means applying one function to the results of another. If we have two functions, say \( f(x) \) and \( g(x) \), then \( (f \circ g)(x) \) means \( f(g(x)) \).

Find \((f \circ g)(x)\)

Given \( f(x) = \log_{2}(x) \) and \( g(x) = 2^{x} \), we need to find \( (f \circ g)(x) = f(g(x)) \). Substituting \( g(x) \) into \( f(x) \), we get:\( (f \circ g)(x) = f(2^{x}) = \log_{2}(2^{x}) \). Since \( \log_{2}(2^{x}) = x \), we have \( (f \circ g)(x) = x \).

Find Domain of \((f \circ g)(x)\)

For \( (f \circ g)(x) \), first check the domains of \( f \) and \( g \). The function \( g(x) = 2^{x} \) is defined for all real \( x \), and the output of \( g \), which is \( 2^x \), will always be positive, fitting the domain of \( f(x) = \log_{2}(x) \) which requires positive input (excluding zero). So, the domain is all real numbers: \( (-\infty, \infty) \).

Find \((g \circ f)(x)\)

Now, \( (g \circ f)(x) = g(f(x)) \). Substituting \( f(x) = \log_{2}(x) \) into \( g(x) \), we get:\( (g \circ f)(x) = g(\log_{2}(x)) = 2^{\log_{2}(x)} \). Recognizing the identity \( 2^{\log_{2}(x)} = x \), we have \( (g \circ f)(x) = x \).

Find Domain of \((g \circ f)(x)\)

For \( (g \circ f)(x) \), first check the domains of \( f \) and \( g \). The function \( f(x) = \log_{2}(x) \) is defined for \( x > 0 \), making the input to \( g \) always valid. Therefore, the domain of \( g(f(x)) \) is \( (0, \infty) \).

Find \((f \circ g)(3)\)

From Step 2, we found \( (f \circ g)(x) = x \), so \( (f \circ g)(3) = 3 \).

Find \((f \circ h)(x)\)

Next, \( (f \circ h)(x) = f(h(x)) \). Given \( h(x) = 4x \), we substitute into \( f \):\( (f \circ h)(x) = f(4x) = \log_{2}(4x) \).

Find Domain of \((f \circ h)(x)\)

For \( (f \circ h)(x) \), consider the domain of \( f \). Since \( f(x) = \log_{2}(x) \), we need \( 4x > 0 \). Solving \( 4x > 0 \) gives \( x > 0 \). Therefore, the domain of \( (f \circ h)(x) \) is \( (0, \infty) \).

Find \((f \circ h)(8)\)

From Step 7, we have \( (f \circ h)(x) = \log_{2}(4x) \). Substituting \( x = 8 \):\( (f \circ h)(8) = \log_{2}(4 \times 8) = \log_{2}(32) = 5 \).

Key concepts

Domain of a Function

The domain of a function represents all the possible input values (x-values) for which the function is defined. For instance, when dealing with logarithmic functions like \( f(x) = \log_{2} x \), it's essential to remember that the input \( x \) must be positive. In simple terms, you cannot take the logarithm of zero or a negative number.

Understanding the domain helps us determine where a function is applicable. For example, the domain of the logarithmic function \( f(x) = \log_{2} (x) \) is \( (0, \infty) \) because of its input restrictions.
Given this, if we have a composite function like \( (f \circ h)(x) \) where \( h(x) = 4x \), we need to ensure that \( 4x \gt 0 \). Solving \( 4x \gt 0 \) gives us \( x \gt 0 \), setting the domain of this composite function to \( (0, \infty) \).

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. If we look at the function \( y = \log_{2} (x) \), it means \( 2^{y} = x \). Here are some key points to help you understand logarithmic functions better:

• The base of the logarithm (2 in \( \log_{2} \)) determines the scale of growth or decay.
• Logarithms can only take positive arguments. Therefore, \( \log_{2} (x) \) is undefined for \( x \leq 0 \).
• The logarithm of 1, regardless of the base, is always 0 because any number raised to the power of 0 is 1.

When we do function compositions involving logarithms, such as \( (f \circ h)(x) \ = \log_{2}(4x) \), we understand that the output of one function becomes the input of another, emphasizing the importance of knowing the domains and behavior of these functions.

Exponential Functions

Exponential functions grow at a rate proportional to their current value. A typical example is \( g(x) = 2^{x} \). Let's break down some properties:

• Exponential functions have a base (in this case, 2) raised to the power of the variable \( x \).
• The function \( 2^{x} \) is defined for all real \( x \).
• It always returns positive values, i.e., \( 2^{x} \gt 0 \) for any real number \( x \).

These properties help us when we compose functions. For example, \( (f \circ g)(x) \), where \( f(x) = \log_{2} x \) and \( g(x) = 2^{x} \), simplifies to \( f(2^{x}) = \log_{2}(2^{x}) = x \). This neat identity occurs because logarithmic and exponential functions are natural inverses of each other.

Composite Functions

Composite functions involve combining two functions where the output of one function becomes the input of another. This is usually denoted as \( (f \circ g)(x) \) which translates to \( f(g(x)) \). Consider the example where \( f(x) = \log_{2} x \) and \( g(x) = 2^{x} \):

• First, we apply \( g(x) \): \( g(x) = 2^{x} \).
• Next, the result of \( g(x) \) becomes the input for \( f \): \( f(g(x)) = f(2^{x}) \).
• Since \( f \) is a logarithm base 2, we get: \( \log_{2}(2^{x}) = x \).

Therefore, \( (f \circ g)(x) \) simplifies to \( x \). Understanding composite functions involves knowing the individual functions and their domains to ensure the operations are valid.