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Chapter 1: Problem 20

Let \(f(x)=2 x^{3}-4 .\) Find a function \(y=g(x)\) so that \((f \circ g)(x)=x+2\).

Short Answer

Expert verified
\(g(x) = \sqrt[3]{\frac{x + 6}{2}}\) satisfies \((f \circ g)(x) = x + 2\).

Step by step solution

01

Understanding the Given Function Composition

We are given that \((f \circ g)(x) = x + 2\). This means when we apply the function \(g(x)\) first, and then \(f(x)\), the final output should be \(x + 2\). Here, \(f(x) = 2x^3 - 4\). We need to find a function \(g(x)\) such that substituting \(g(x)\) into \(f(x)\) gives us \(x + 2\): \(f(g(x)) = x + 2\).
02

Setting Up the Equation

Since \(f(g(x)) = 2(g(x))^3 - 4\), we set this equal to \(x + 2\):\[2(g(x))^3 - 4 = x + 2\].
03

Solving for g(x)

To solve for \(g(x)\), first isolate \(g(x)\) on one side of the equation by adding 4 to both sides:\[2(g(x))^3 = x + 6\].Next, divide both sides by 2:\[(g(x))^3 = \frac{x + 6}{2}\].Finally, take the cube root of both sides:\[g(x) = \sqrt[3]{\frac{x + 6}{2}}\].
04

Verification

To verify, substitute \(g(x) = \sqrt[3]{\frac{x + 6}{2}}\) back into \(f\) and check if \(f(g(x)) = x + 2\):Start with \(f(x) = 2x^3 - 4\) and \(g(x) = \sqrt[3]{\frac{x + 6}{2}}\):\[f(g(x)) = 2\left(\sqrt[3]{\frac{x + 6}{2}}\right)^3 - 4\].Simplifying, we get:\[f(g(x)) = 2\left(\frac{x + 6}{2}\right) - 4\].This equals \((x + 6) - 4 = x + 2\).The function \(g(x)\) is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Functions
Inverse functions are essential for understanding how to reverse the effect of a function and accomplish its opposite action.
  • Think of a function as a machine that takes an input, performs a set of operations, and delivers an output.
  • The inverse function does the reverse; it takes that output and returns it to the original input.
In the exercise, we are tasked with finding a function \(g(x)\) that operates as an inverse to part of the composition \((f \circ g)(x) = x + 2\). This implies that when we apply \(f\) after \(g\), it "undoes" part of these effects, making the expression simplify to \(x + 2\). The point of understanding inverses is crucial in solving the exercise, as \(g(x)\) must be "carefully crafted" to allow \(f(g(x))\) to reverse-engineer to \(x + 2\), making them suitable partners in function composition.
Algebraic Manipulation
Algebraic manipulation involves a series of logical steps to transform equations or expressions to a more useful form. It's like a toolkit that helps you simplify, rearrange, or solve for unknowns. In our solution process:
  • Start with the equation \(2(g(x))^3 - 4 = x + 2\) and add 4 to both sides to balance the operation, resulting in \(2(g(x))^3 = x + 6\).
  • Then, isolate \(g(x)^3\) by dividing both sides by 2, giving you \((g(x))^3 = \frac{x + 6}{2}\).
  • Finally, extract the cube root to solve for \(g(x)\): \(g(x) = \sqrt[3]{\frac{x + 6}{2}}\).
Each move in this sequence is designed to untangle \(g(x)\) step-by-step, reaching the function that meets the desired composition criteria. Mastery of algebraic manipulation empowers problem-solving in both simple and complex mathematical challenges.
Cube Roots
Understanding cube roots is key in certain mathematical relationships, especially when dealing with polynomials like cubes of variables. A cube root finds the number that, when multiplied by itself twice, results in the original number.For instance, the cube root of 8 is 2, because \(2 \times 2 \times 2 = 8\).In the context of the exercise, once you reached \((g(x))^3 = \frac{x + 6}{2}\), it was necessary to find the cube root to solve for \(g(x)\). Here鈥檚 the thought process:
  • There's a relationship between cubes and cube roots, similar to how squares relate to square roots.
  • The operation \(g(x) = \sqrt[3]{\frac{x + 6}{2}}\) undoes the cubing observed in \((g(x))^3\), directly finding \(g(x)\).
Recognizing this inverse operation at this stage highlights how cube roots are invaluable for unlocking solutions.

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