Problem 26 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\frac{2 x}{x+3}\) and \(g(x)=\frac{1}{x+1}\). (a) Find \(f(g(2))\). (b) Find \(f(g(x))\) and simplify your answer. Be sure that your answer is in agreement with the concrete case from part (a).

Step-by-Step Solution

Answer

The output of \(f(g(2))\) is 0.2 and the simplified form of \(f(g(x))\) is \(\frac{2}{4+3x}\).

Step 1: Find the value of \(g(2)\)

First, by substituting \(x = 2\) into \(g(x) = \frac{1}{x+1}\), we get:\(g(2) = \frac{1}{2+1} = \frac{1}{3}\).

Step 2: Compute for \(f(g(2))\)

Now, replace \(x\) in \(f(x)\) by \(g(2)\):\(f(g(2)) = f(\frac{1}{3}) = \frac{2 * \frac{1}{3}}{\frac{1}{3} + 3} = \frac{2/3}{10/3} = \frac{2}{10} = 0.2\).

Step 3: Compute for \(f(g(x))\)

Next, replace \(x\) in \(f(x)\) by \(g(x)\):\(f(g(x)) = f\left(\frac{1}{x+1}\right) = \frac{2 * \frac{1}{x+1}}{\frac{1}{x+1} + 3}\).

Step 4: Simplify \(f(g(x))\)

Simplify further to:\(f(g(x)) = \frac{2/(x+1)}{1/(x+1) + 3} = \frac{2}{1+3(x+1)} = \frac{2}{1+3x+3} = \frac{2}{4+3x}\).

Key Concepts

Function Composition

When we talk about function composition, we are referring to the process of applying one function to the result of another function. If you have functions labeled as f(x) and g(x), their composition is written as f(g(x)), which means you first apply g to x, and then apply f to the result of g(x). Imagine this as a 'function machine' in which g(x) transforms the initial input, and then f(x) further transforms that output.

Function composition isn't just about plugging one equation into another—it requires a careful handling of the resulting expressions to ensure accuracy. The order in which the functions are composed matters a lot, as f(g(x)) can yield a completely different result from g(f(x)). It's essential to follow the correct order of operations to avoid any confusion or errors when working with compositions of functions.

Simplifying Complex Fractions

Complex fractions can look intimidating, but they follow the same principles of simplification as simpler fractions. A complex fraction has a fraction in the numerator, denominator, or both. The key to simplifying them is to find a common denominator and to rewrite the entire expression as a single fraction.

One effective method for simplifying complex fractions is the multiplication of the numerator and the denominator by the least common denominator (LCD) of all the fractions involved. This will help 'clear' the fractions within the fractions. Remember that in the world of algebra, simplification is about making expressions as straightforward as possible, whether that means factoring, canceling out terms, or finding more concise ways to represent the information. The goal is to reduce the complexity while maintaining mathematical integrity.

Function Evaluation

Evaluating a function means that you're finding the value of the function for a particular value of x. This involves substituting the given number or expression for x into the function and simplifying to find the result. It's also a fundamental step in function composition, where you evaluate one function, and then plug that result into another function.

Proper function evaluation is crucial in calculus and other higher math, as it forms the basis for creating accurate graphs, solving equations, and understanding the behavior of functions. To evaluate functions correctly, you must follow the order of operations, carefully execute any required algebra, and simplify your results when possible. This process not only helps in understanding the specific value of a function at a certain point but also in understanding how functions operate in general.