Problem 10 - Thomas Calculus

Question

In Exercises \(7-10,\) write a formula for \(f \circ g \circ h\) $$ f(x)=\frac{x+2}{3-x}, \quad g(x)=\frac{x^{2}}{x^{2}+1}, \quad h(x)=\sqrt{2-x} $$

Step-by-Step Solution

Answer

The formula for \(f \circ g \circ h\) is \(\frac{8-3x}{7-2x}\).

Step 1: Understand Function Composition

The problem requires finding a formula for the composition of three functions, specifically \((f \circ g \circ h)(x)\). This means we need to find \(f(g(h(x)))\). We must apply each function one at a time, starting with the innermost function and working outward.

Step 2: Apply the Innermost Function

Start with function \(h(x) = \sqrt{2-x}\). This is the innermost function, so we substitute this expression wherever \(x\) appears in \(g(x)\). Thus, \(g(h(x)) = g(\sqrt{2-x})\).

Step 3: Substitute Into Second Function

Substitute \(h(x)\) into \(g(x)\). With \(g(x) = \frac{x^{2}}{x^{2}+1}\), replace \(x\) with \(\sqrt{2-x}\):\[ g(h(x)) = \frac{(\sqrt{2-x})^2}{(\sqrt{2-x})^2 + 1} \]Simplify the expression:\[ g(h(x)) = \frac{2-x}{2-x+1} = \frac{2-x}{3-x} \]

Step 4: Substitute Into the Outermost Function

Now, substitute \(g(h(x))\) into \(f(x)\). With \(f(x) = \frac{x+2}{3-x}\), we replace \(x\) with \(\frac{2-x}{3-x}\):\[ f(g(h(x))) = \frac{\frac{2-x}{3-x} + 2}{3 - \frac{2-x}{3-x}} \]

Step 5: Simplify the Expression

Simplify the expression from the previous step:First, simplify the numerator:\[ \frac{2-x}{3-x} + 2 = \frac{2-x + 2(3-x)}{3-x} = \frac{2 - x + 6 - 2x}{3-x} = \frac{8 - 3x}{3-x} \]Now, simplify the denominator:\[ 3 - \frac{2-x}{3-x} = \frac{3(3-x) - (2-x)}{3-x} = \frac{9 - 3x - 2 + x}{3-x} = \frac{7 - 2x}{3-x} \]Combine them together:\[ f(g(h(x))) = \frac{\frac{8-3x}{3-x}}{\frac{7-2x}{3-x}} = \frac{8-3x}{7-2x} \]

Step 6: Result

The final formula for \(f \circ g \circ h\) is obtained by plugging \(g(h(x))\) into \(f(x)\) and simplifying:\[ (f \circ g \circ h)(x) = \frac{8-3x}{7-2x} \]

Key Concepts

Composition of Functions

Function composition is a crucial concept in mathematics that involves applying one function to the results of another. It's often shown using the notation \((f \circ g \circ h)(x)\), meaning you first apply the function \(h(x)\), then \(g(x)\), and finally \(f(x)\). This process is akin to following a neat sequence: starting from the innermost function and working your way outwards.

To get your head around this, think of it as a series of machines. Each machine (function) takes in an input, does its own specific operation on it, and then passes the result to the next machine. Starting with \(h(x) = \sqrt{2-x}\), which transforms \(x\) by taking the square root of \(2-x\). The output of \(h\) becomes the input for \(g\), which is \(g(x) = \frac{x^2}{x^2+1}\). Finally, \(f(x) = \frac{x+2}{3-x}\) processes the result from \(g\) to give the final outcome.

• The order matters: always start from the innermost function.
• Your result from one function becomes the input to the next.

Understanding this sequence helps in tackling complex algebraic problems by breaking them down into simpler, manageable parts, and also lays a foundation for advanced topics in calculus, such as derivative chain rules.

Simplifying Expressions

Simplifying expressions is at the heart of resolving compositions. Each time you determine the output of one function and substitute it into the next, simplify the expression as much as possible. This is important because it makes handling and understanding the problem easier.

Take for instance the step from \(g(h(x)) = \frac{\left(\sqrt{2-x}\right)^2}{\left(\sqrt{2-x}\right)^2 + 1}\). Squaring \(\sqrt{2-x}\) gives \(2-x\), simplifying the fraction to \(\frac{2-x}{3-x}\). After substituting \(g(h(x))\) into \(f(x)\), you are left with the complex fraction:
\[ \frac{\frac{8-3x}{3-x}}{\frac{7-2x}{3-x}} \] which further simplifies to \(\frac{8-3x}{7-2x}\).

• Simplification reduces complexity.
• It prevents errors further along in problem-solving.
• It's the key to cleaner, concise results.

This step is crucial in mathematics as it allows for working with expressions that are more understandable and manageable, ultimately leading to the neat final functions.

Nested Functions

Nested functions refer to situations where functions are placed within other functions, much like nested dolls. This idea is central to understanding function composition, as each function's output becomes the next function's input.

For our example, it started with the innermost function, \(h(x) = \sqrt{2-x}\), and nested it within \(g(x) = \frac{x^2}{x^2+1}\). Then, \(g(h(x))\) itself was used as an input for \(f(x) = \frac{x+2}{3-x}\).

• Nesting requires increasingly careful substitutions.
• Focus on systematically replacing variables step by step.

The beauty of mastering nested functions lies in instilling a deeper understanding of how functions interact. It fosters an ability to break down potentially intimidating mathematics into bits that are comprehensible, making challenging problems much simpler to solve.