Problem 9 - Precalculus

Question

Use the given pair of functions to find the following values if they exist. $$\bullet (g \circ f)(0)$$ $$\bullet (f \circ g)(-1)$$ $$\bullet (f \circ f)(2)$$ $$\bullet (g \circ f)(-3)$$ $$\bullet (f \circ g)\left(\frac{1}{2}\right)$$ $$\bullet (f \circ f)(-2)$$ $$f(x)=\frac{3}{1-x}, g(x)=\frac{4 x}{x^{2}+1}$$

Step-by-Step Solution

Answer

1. \(\frac{6}{5}\); 2. 1; 3. \(\frac{3}{4}\); 4. \(\frac{48}{25}\); 5. -5; 6. Undefined

Step 1: Understanding Composite Functions

Composite functions involve applying one function to the result of another function. For example, \( (g \circ f)(x) \) means you first apply \( f(x) \) then apply \( g(x) \) to the result, and \( (f \circ g)(x) \) means you apply \( g(x) \) first and then \( f(x) \).

Step 2: Calculating (g o f)(0)

To compute \( (g \circ f)(0) \), substitute \( x = 0 \) into \( f(x) = \frac{3}{1-x} \), which gives \( f(0) = \frac{3}{1} = 3 \). Then substitute \( f(0) = 3 \) into \( g(x) = \frac{4x}{x^2+1} \), giving \( g(3) = \frac{4 \cdot 3}{3^2 + 1} = \frac{12}{10} = \frac{6}{5} \).

Step 3: Calculating (f o g)(-1)

To find \( (f \circ g)(-1) \), compute \( g(-1) = \frac{4(-1)}{(-1)^2 + 1} = \frac{-4}{2} = -2 \), and then use this result in \( f(x) \): \( f(-2) = \frac{3}{1 - (-2)} = \frac{3}{3} = 1 \).

Step 4: Calculating (f o f)(2)

For \( (f \circ f)(2) \), find \( f(2) = \frac{3}{1 - 2} = \frac{3}{-1} = -3 \). Then substitute back into \( f(x) \) to get \( f(-3) = \frac{3}{1 - (-3)} = \frac{3}{4} \).

Step 5: Calculating (g o f)(-3)

Compute \( f(-3) = \frac{3}{1 - (-3)} = \frac{3}{4} \), then use this result in \( g(x) \): \( g\left(\frac{3}{4}\right) = \frac{4 \times \frac{3}{4}}{\left(\frac{3}{4}\right)^2 + 1} = \frac{3}{\frac{9}{16} + 1} = \frac{3}{\frac{25}{16}} = \frac{48}{25} \).

Step 6: Calculating (f o g)(1/2)

First, find \( g\left(\frac{1}{2}\right) = \frac{4\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)^2 + 1} = \frac{2}{\frac{1}{4} + 1} = \frac{2}{\frac{5}{4}} = \frac{8}{5} \). Then input \( \frac{8}{5} \) into \( f(x) \) so that \( f\left(\frac{8}{5}\right) = \frac{3}{1 - \frac{8}{5}} = \frac{3}{-\frac{3}{5}} = -5 \).

Step 7: Calculating (f o f)(-2)

Calculate \( f(-2) = \frac{3}{1 - (-2)} = \frac{3}{3} = 1 \), and substitute back into \( f(x) \): \( f(1) = \frac{3}{1-1} \), which is undefined because dividing by zero is not allowed.

Key Concepts

Understanding Function Composition

Function composition is like a "function of a function" where you apply one function to the results of another. It's essential to understand this concept when solving composite function problems, such as \((g \circ f)(x)\) or \((f \circ g)(x)\). Understanding the sequence is key:

• For \((g \circ f)(x)\), you first apply \(f(x)\), then apply \(g(x)\) to the result.
• For \((f \circ g)(x)\), start with \(g(x)\) and then use \(f(x)\) on its output.

Imagine it like a pipeline where data is transformed as it moves from function to function.
Start with an initial value, transform it with the first function, then take that result and input it into the second function. Each function impacts the journey of your initial data point.

Exploring Domain and Range in Composite Functions

The domain and range are crucial when dealing with composite functions. They tell us what values can go into the function (domain) and what can come out (range).
Understanding these restrictions ensures accurate calculations and avoidance of undefined results. The domain of a composite function, say \((g \circ f)(x)\), depends on:

• The domain of \(f\), since \(f\) must produce values that \(g\) can accept.
• The outputs of \(f\) must fall within the domain of \(g\).

Consider the function \(f(x) = \frac{3}{1-x}\). Its domain excludes \(x = 1\), where the denominator becomes zero, leading to undefined results.
For \((f \circ f)(x)\), check if the result from the first application of \(f\) falls in the domain of the second \(f\).
Ensuring each function's output matches the next input maintains the integrity of the function chain.

Getting the Hang of Function Evaluation

Function evaluation is the process of substituting specific values into a function and calculating the output. It forms the core of understanding composite functions. Directly substituting values can sometimes create zero denominators or lead to undefined results.
Let's simplify the earlier example:

• To evaluate \((g \circ f)(0)\), substitute \(0\) into \(f(x)\), then take that result and input it into \(g(x)\).
• This step-by-step substitution helps simplify complex expressions.

Always start with the innermost function when dealing with composites and use precise calculations.
If an operation results in dividing by zero, the function is undefined for that value.
Evaluate methodically, ensuring each outcome fits into the next calculation. This process provides clear outputs and resolves intricate expressions one step at a time.