Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(g(x)=x^{2}+1\), find formulas for \(g^{3}(x)\) and \((g \circ g \circ g)(x) .\)

Step-by-Step Solution

Answer

\(g^3(x) = x^6 + 3x^4 + 3x^2 + 1\); \((g \circ g \circ g)(x) = x^8 + 4x^6 + 16x^4 + 8x^2 + 5\).

Step 1: Find the formula for \(g^3(x)\)

The notation \(g^3(x)\) means \( (g(x))^3 \). So, we need to cube the expression for \(g(x)\). Since \(g(x) = x^2 + 1\), we have:\[ g^3(x) = (x^2 + 1)^3 \]

Step 2: Expand \((g(x))^3\)

To compute \((x^2 + 1)^3\), we can use the binomial theorem:\[ (x^2 + 1)^3 = (x^2)^3 + 3(x^2)^2(1) + 3(x^2)(1)^2 + 1^3 \]Evaluate each term:- \((x^2)^3 = x^6\)- \(3(x^2)^2(1) = 3x^4\)- \(3(x^2)(1) = 3x^2\)- \(1^3 = 1\)Combine all terms:\[ g^3(x) = x^6 + 3x^4 + 3x^2 + 1 \]

Step 3: Find the formula for \((g \circ g \circ g)(x)\)

The notation \((g \circ g \circ g)(x)\) denotes the function \(g\) applied three times. Start with \(g(g(x))\), where \(g(x) = x^2 + 1\):1. Compute \(g(g(x))\): - Replace \(x\) in \(g(x)\) by \(g(x)\): \[ g(g(x)) = g(x^2 + 1) = (x^2 + 1)^2 + 1 \]2. Simplify \((x^2 + 1)^2 + 1\): - \((x^2 + 1)^2 = x^4 + 2x^2 + 1\) - Add 1: \(x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2\)3. Compute \(g(g(g(x)))\): - Replace \(x\) in \(g(x)\) by \(g(g(x))\): \[ g(g(g(x))) = g(x^4 + 2x^2 + 2) \] - This becomes: \[ (x^4 + 2x^2 + 2)^2 + 1 \]

Step 4: Simplify \((x^4 + 2x^2 + 2)^2 + 1\)

Use the binomial theorem to expand:\[ (x^4 + 2x^2 + 2)^2 = (x^4)^2 + 2(x^4)(2x^2) + 2(2x^2)^2 + 2(2x^2)(2x^2) + 2^2 \]Simplify each term:- \((x^4)^2 = x^8\)- \(2(x^4)(2x^2) = 4x^6\)- \(2(2x^2)^2 = 8x^4\)- \(2(2x^2)(2x^2) = 8x^4\)- \(2^2 = 4\)So, by combining, we get:\[ (x^4 + 2x^2 + 2)^2 = x^8 + 4x^6 + 16x^4 + 8x^2 + 4 \] Add 1 to get \(g(g(g(x)))\):\[ (g \circ g \circ g)(x) = x^8 + 4x^6 + 16x^4 + 8x^2 + 5 \]

Key Concepts

Binomial Theorem

The binomial theorem is a powerful mathematical tool used to expand expressions that are raised to a power. It helps in breaking down binomial expressions like \((a + b)^n\) into a sum of terms involving powers of \(a\) and \(b\).
The basic idea is to use the formula:\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]where \(\binom{n}{k}\) are the binomial coefficients, found in Pascal's triangle. Each term in the expansion corresponds to a specific value of \(k\).
In our exercise, we used the binomial theorem to expand \((x^2 + 1)^3\) and later \((x^4 + 2x^2 + 2)^2\).

• For the expansion \((x^2 + 1)^3\), we calculated individual terms like \((x^2)^3\), \(3(x^2)^2(1)\), \(3(x^2)(1)^2\), and \(1^3\).
• Similarly, for \((x^4 + 2x^2 + 2)^2\), each term was expanded using the sum of different products of powers of \(x^4\), \(2x^2\), and \(2\).

Understanding the binomial theorem not only makes polynomial expansion simpler but also paves the way for tackling more complex expressions.

Nested Functions

Nested functions occur when a function is applied several times within itself, like \((g \circ g \circ g)(x)\), denoting the application of \(g(x)\) three times consecutively.
When dealing with nested functions, the key is to substitute each instance of \(x\) with the previous result.Here's how it works in this exercise:

• Start with \(g(x) = x^2 + 1\).
• First apply the function to itself: compute \(g(g(x)) = (x^2 + 1)^2 + 1\).
• Next, apply the function again: \(g(g(g(x)))\) changes \(x\) into \(g(g(x))\), resulting in \((x^4 + 2x^2 + 2)^2 + 1\).

Understanding nested functions is crucial for recognizing how iterations and recursive rules work in various mathematical contexts, leading to complex and interesting functions as results.

Polynomial Expansion

Polynomial expansion involves expressing a polynomial raised to a given power as a sum of terms, each of which is a product of constants and powers of variables. This concept is closely tied with the binomial theorem.
When expanding polynomials, each new term in the expanded form results from observing combinations of terms from the original polynomial raised to different powers.
Take for example: \((x^2 + 1)^3\)

• This expansion involves combining terms derived from multiplying out \((x^2 + 1)\) across each layer, growing more terms as the power increases.

Each increase in power adds more complexity and more terms, like:

• The \(x^6\) term from \((x^2)^3\),
• or the combination of x terms and constant multiplications producing the final output of six terms in total.

In our example, we expanded polynomials to find expressions for powers of \(g(x)\) and results when functions are composed repeatedly, using a step-by-step process to make sense of the extended complexity.