Problem 16 - Calculus for Biology and Medicine

Question

Suppose that \(f(x)=\frac{1}{x+1}, x \neq-1\), and \(g(x)=2 x^{2}, x \in \mathbf{R}\). (a) Find \((f \circ g)(x)\). (b) Find \((g \circ f)(x)\). In both (a) and (b), find the domain.

Step-by-Step Solution

Answer

(f ◦ g)(x) = \frac{1}{2x^2 + 1}, domain: \mathbb{R}. (g ◦ f)(x) = \frac{2}{(x+1)^2}, domain: \mathbb{R} \backslash \{-1\}.

Step 1: Understand Composition of Functions (f ◦ g)(x)

When asked to find \((f \circ g)(x)\), it means we want to substitute the function \(g(x)\) into \(f(x)\). Thus, \((f \circ g)(x) = f(g(x))\). So, you'll replace \(x\) in \(f(x) = \frac{1}{x+1}\) with \(g(x) = 2x^2\).

Step 2: Substitute g(x) into f(x)

Substitute \(g(x)\) into \(f(x)\):\[f(g(x)) = f(2x^2) = \frac{1}{2x^2 + 1}\]

Step 3: Determine the Domain of (f ◦ g)(x)

The domain of \((f \circ g)(x)\) is all real numbers where the expression \(2x^2 + 1 eq 0\) is defined. Since \(2x^2 + 1 > 0\) for all real \(x\), there's no restriction from the function itself. Thus, the domain is all real numbers \(\mathbb{R}\).

Step 4: Understand Composition of Functions (g ◦ f)(x)

When asked to find \((g \circ f)(x)\), it means we want to substitute the function \(f(x)\) into \(g(x)\). Thus, \((g \circ f)(x) = g(f(x))\). So, you'll replace \(x\) in \(g(x) = 2x^2\) with \(f(x) = \frac{1}{x+1}\).

Step 5: Substitute f(x) into g(x)

Substitute \(f(x)\) into \(g(x)\):\[g(f(x)) = g\left(\frac{1}{x+1}\right) = 2\left(\frac{1}{x+1}\right)^2 = \frac{2}{(x+1)^2}\]

Step 6: Determine the Domain of (g ◦ f)(x)

The domain of \((g \circ f)(x)\) is determined by both functions. From \(f(x)\), we know \(x eq -1\). Additionally, the expression is defined wherever \((x+1)^2 eq 0\), which also requires \(x eq -1\). Therefore, the domain is all real numbers except \(-1\), i.e., \(x \in \mathbb{R}\) and \(x eq -1\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (commonly represented as \(x\)) that the function can accept without causing errors like division by zero. For example, if we have a function \(f(x) = \frac{1}{x+1}\), we must ensure that the denominator, \(x+1\), is never zero as this would make the expression undefined. Hence, we exclude \(x = -1\) from the domain. When determining the domain, always look for values that result in the denominator being zero or any other condition that might render the function undefined, such as negative values under even roots. This concept is key when dealing with rational functions and composite functions.
To find the domain:

• Identify terms where division by zero could occur.
• Exclude these values from the domain.
• Consider any other conditions (like square roots) that may affect the domain.

Composite Function Calculation

Composite function calculation involves substituting one function into another. We denote this by \((f \circ g)(x)\), meaning to substitute \(g(x)\) into \(f(x)\), resulting in \(f(g(x))\). Taking the original functions from our exercise:

• For \((f \circ g)(x)\), substitute \(g(x) = 2x^2\) into \(f(x) = \frac{1}{x+1}\), resulting in \(f(2x^2) = \frac{1}{2x^2 + 1}\).

Next, calculate domains: since \(2x^2 + 1 eq 0\) for all real \(x\), every real \(x\) is valid. Thus, it covers the entire set of real numbers, \(\mathbb{R}\).
Next, consider \((g \circ f)(x)\):

• Substitute \(f(x) = \frac{1}{x+1}\) into \(g(x) = 2x^2\), leading to \(g\left(\frac{1}{x+1}\right) = \frac{2}{(x+1)^2}\).

In this case, the domain comprises all real numbers except \(x = -1\), where the output is undefined.

Rational Functions

Rational functions are expressions that involve the ratio of two polynomials. Consider the general form \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not zero. These types of functions are frequently encountered in algebra and calculus because they present intriguing behavior near the values that make the denominator zero.
Analyzing the behavior:

• For \(f(x) = \frac{1}{x+1}\): It is undefined at \(x = -1\) because the denominator becomes zero.
• Avoid points of division by zero by understanding and calculating its domain.
• Simplify wherever possible to make calculations easier when combining or composing with other functions.

Rational functions will often represent real-world scenarios like rates or proportions, making understanding their domain and properties valuable for both theoretical and applied contexts.