Problem 42 - Algebra and Trigonometry

Question

\(41-44\) Find \(f \circ g \circ h\) $$ f(x)=\frac{1}{x}, \quad g(x)=x^{3}, \quad h(x)=x^{2}+2 $$

Step-by-Step Solution

Answer

The composition \(f \circ g \circ h (x) = \frac{1}{x^6 + 6x^4 + 12x^2 + 8}\).

Step 1: Understand Composition of Functions

When we are asked to find \((f \circ g \circ h)(x)\), it means we need to evaluate the composition of three functions, which is equivalent to \(f(g(h(x)))\). We first find \(h(x)\), then use the result to find \(g(h(x))\), and finally use this to find \(f(g(h(x)))\).

Step 2: Evaluate h(x)

The function \(h(x)\) is given as \(x^2+2\). So, when we evaluate \(h(x)\), we simply substitute any given \(x\) into this expression, resulting in \(h(x) = x^2 + 2\).

Step 3: Plug h(x) into g(x)

Next, we need to find \(g(h(x))\), where \(g(x) = x^3\). Substitute \(h(x)\) for \(x\) in \(g(x)\), resulting in \(g(h(x)) = (x^2 + 2)^3\).

Step 4: Simplify g(h(x))

We expand \((x^2 + 2)^3\) to simplify it further. Using binomial expansion, \((x^2 + 2)^3 = x^6 + 6x^4 + 12x^2 + 8\).

Step 5: Plug g(h(x)) into f(x)

Finally, we compute \(f(g(h(x)))\), where \(f(x) = \frac{1}{x}\). Substitute \(g(h(x))\) into \(f(x)\) to get \(f(g(h(x))) = \frac{1}{x^6 + 6x^4 + 12x^2 + 8}\).

Key Concepts

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input value. It involves substituting the input value for the variable in the function and performing any operations necessary to find the result. In our exercise, the functions we are dealing with are \( f(x) = \frac{1}{x} \), \( g(x) = x^3 \), and \( h(x) = x^2 + 2 \). To evaluate these:

• For \( h(x) \), replace \( x \) with the given value: if \( x = 1 \), then \( h(1) = 1^2 + 2 = 3 \).
• For \( g(x) \), substitute the value, say \( x = 2 \), to get \( g(2) = 2^3 = 8 \).
• For \( f(x) \), if \( x = 4 \), then \( f(4) = \frac{1}{4} \).

Each function takes an \( x \) value, performs particular operations, and provides an output. This process of calculation demonstrates how each individual function behaves for the input provided.

Binomial Expansion

Binomial expansion is a technique used to expand expressions that are raised to a power. This is especially useful for expressions of the form \((a + b)^n\). The expansion makes use of the binomial theorem, which states\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]where \( \binom{n}{k} \) is the binomial coefficient. When we apply binomial expansion to expressions like \( (x^2 + 2)^3 \), we use the formula to expand:

• Start with \( x^{2*3} = x^6 \).
• Add the next term: \( 3 \times (x^2)^2 \times 2 = 6x^4 \).
• Next term: \( 3 \times x^2 \times 2^2 = 12x^2 \).
• Follow with the final constant term: \( 2^3 = 8 \).

So, \( (x^2 + 2)^3 \) becomes \( x^6 + 6x^4 + 12x^2 + 8 \). This process allows you to treat complex power expressions in a manageable way.

Function Substitution

Function substitution is a powerful tool that involves replacing one function within another. This process is essential for evaluating complex compositions of functions, like \((f \circ g \circ h)(x)\). By substituting functions step-by-step, we simplify the composition:

• First, calculate \( h(x) \) and use it as the input for \( g(x) \): find \( g(h(x)) \), which is \( (x^2 + 2)^3 \).
• Next, take the result from \( g(h(x)) \) and use it in \( f(x) \): substitute into \( f(x) = \frac{1}{x} \) to get \( f(g(h(x))) = \frac{1}{x^6 + 6x^4 + 12x^2 + 8} \).

This systematic approach not only simplifies the computation but also provides a deeper understanding of how functions interact with each other. By substituting functions appropriately, you can unravel the complexity of nested compositions effectively.