Problem 16 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = \frac{1}{2x^2+1}\), domain: \(x \in \mathbb{R}\). (b) \((g \circ f)(x) = \frac{2}{(x+1)^2}\), domain: \(x \in \mathbb{R}, x \neq -1\).

Step 1: Understanding Function Composition

Function composition \((f \circ g)(x)\) means applying function \(g\) first and then applying function \(f\) to the result. Function composition \((g \circ f)(x)\) means applying function \(f\) first and then applying function \(g\) to the result.

Step 1: Find \((f \circ g)(x)\)

To solve \((f \circ g)(x)\), substitute \(g(x)\) into \(f(x)\). Thus, \((f \circ g)(x) = f(g(x)) = f(2x^2)\). Substituting \(2x^2\) into \(f(x)\), we get:\[f(2x^2) = \frac{1}{2x^2 + 1}\].

Step 2: Domain of \((f \circ g)(x)\)

The domain of a composite function \((f \circ g)(x)\) requires that the output of \(g(x)\), which is \(2x^2\), must not result in an expression where \(f(x)\) is undefined. Since \(f(x)\) is undefined when its denominator equals zero, we find when \(2x^2 + 1 = 0\). However, \(2x^2 + 1\) is always positive, hence \(f(g(x)) = \frac{1}{2x^2 + 1}\) is defined for all \(x\), i.e., the domain of \((f \circ g)(x)\) is \(x \in \mathbb{R}\).

Step 3: Find \((g \circ f)(x)\)

To solve \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). Therefore, \((g \circ f)(x) = g(f(x)) = g\left(\frac{1}{x+1}\right)\).Substituting \(\frac{1}{x+1}\) into \(g(x)\), we compute:\[g\left(\frac{1}{x+1}\right) = 2\left(\frac{1}{x+1}\right)^2 = \frac{2}{(x+1)^2}\].

Step 4: Domain of \((g \circ f)(x)\)

The domain of \((g \circ f)(x)\) consists of values for \(x\) where both \(f(x)\) and \(g(f(x))\) are defined. For \(f(x)\), \(x eq -1\), as that would make the denominator zero. Therefore, the domain of \((g \circ f)(x)\) is \(x \in \mathbb{R}, x eq -1\).

Key Concepts

Understanding Domain of Functions

The domain of a function is the set of all possible input values (commonly represented as x-values) that allow the function to produce a real output without leading to undefined situations. Let's break it down:
- For many functions, a common restriction arises when the function involves division by zero. This occurs in rational functions, where any zero in the denominator results in the value being undefined.
- Additionally, when dealing with composite functions like \(f \circ g\)(x), we must consider each function in the composition individually. We ensure that once the initial function (g(x)) is applied, applying the subsequent function (f(x)) does not lead to an undefined result.
If both these conditions are satisfied, the domain of a composite function can be determined by analyzing the individual functions.
To solve for the domain, you often need to identify the x-values that make the denominator of a rational function equal to zero and exclude them from the domain.

Exploring Composite Functions

Composite functions involve the combination of two functions where the output of one function becomes the input of another. This concept is expressed as \(f \circ g\)(x), which is read as "f composed with g." Here, g(x) is first evaluated, then its result is passed through f(x). Let's simplify it:
- Start with the innermost function. Compute \(g(x)\). For example, given \(g(x) = 2x^2\), you would first find \(2x^2\).
- Then use that outcome in the outer function \(f(x)\). For example, if \(f(x) = \frac{1}{x+1}\), you replace x with the result from \(g(x)\), such as putting \(2x^2\) into f, leading to \(\frac{1}{2x^2 + 1}\).
In a similar fashion, \(g \circ f\)(x) means applying f first and then g. These operations form the basis of manipulating functions in various mathematical contexts.

Understanding Rational Functions and Their Importance

Rational functions are fractions where both the numerator and the denominator are polynomials. An example is \(f(x) = \frac{1}{x+1}\). Rational functions are known for having specific characteristics that impact their domain and behavior:
- The main restriction is any x-value that causes the denominator to be zero, making the function undefined at those points.
- These functions often have vertical asymptotes, which are lines where the value of the function tends to infinity. Exploring these can give insights into the behavior of the function around certain points.
In studying rational functions, you encounter core algebraic techniques such as factorization, long division of polynomials, and identifying asymptotic behavior. Understanding these characteristics of rational functions aids in anticipating their behavior and assessing both their domains and ranges efficiently.