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Chapter 5: Problem 30

Exercises \(11-30:\) Use \(f(x)\) and \(g(x)\) to find a formula for each expression. Identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \((f / g)(x)\) $$ f(x)=x^{2 / 3}-2 x^{1 / 3}+1, \quad g(x)=x^{1 / 3}-1 $$

Short Answer

Expert verified
The expressions are: (a) \((f+g)(x) = x^{2/3} - x^{1/3}\), (b) \((f-g)(x) = x^{2/3} - 3x^{1/3} + 2\), (c) \((fg)(x) = x - 3x^{2/3} + 3x^{1/3} - 1\), (d) \((f/g)(x) = \frac{x^{2/3} - 2x^{1/3} + 1}{x^{1/3} - 1}\). Domains are (a) and (b) all real numbers, (c) all real numbers, (d) all real numbers except \(x=1\).

Step by step solution

01

Understand the Problem

We need to find expressions for \((f+g)(x)\), \((f-g)(x)\), \((fg)(x)\), and \((f/g)(x)\) using the given functions \(f(x)=x^{2/3}-2x^{1/3}+1\) and \(g(x)=x^{1/3}-1\). We also need to determine the domain of each result.
02

Calculate (f+g)(x)

To find \((f+g)(x)\), we sum the two functions:\[(f+g)(x) = f(x) + g(x) = (x^{2/3} - 2x^{1/3} + 1) + (x^{1/3} - 1)\]Simplifying, we get:\[(f+g)(x) = x^{2/3} - x^{1/3}\]
03

Calculate the domain of (f+g)(x)

The domain of \(f(x)\) is all real numbers because the cube root and square root of \(x\) are defined for all real \(x\). Similarly, the domain of \(g(x)\) is all real numbers. Therefore, the domain of \((f+g)(x)\) is all real numbers \( (-\infty, \infty) \).
04

Calculate (f-g)(x)

To find \((f-g)(x)\), subtract \(g(x)\) from \(f(x)\):\[(f-g)(x) = f(x) - g(x) = (x^{2/3} - 2x^{1/3} + 1) - (x^{1/3} - 1)\]Simplifying, we get:\[(f-g)(x) = x^{2/3} - 3x^{1/3} + 2\]
05

Calculate the domain of (f-g)(x)

Similar to \((f+g)(x)\), the domains of \(f(x)\) and \(g(x)\) allow for all real numbers. Thus, the domain of \((f-g)(x)\) is \((-\infty, \infty)\).
06

Calculate (fg)(x)

To find \((fg)(x)\), multiply the functions:\[(fg)(x) = f(x) \cdot g(x) = (x^{2/3} - 2x^{1/3} + 1)(x^{1/3} - 1)\]Expanding this product, we get:\[fg(x) = x^{2/3} \cdot x^{1/3} - x^{2/3} - 2x^{1/3} \cdot x^{1/3} + 2x^{1/3} + x^{1/3} - 1\]Which simplifies to:\[fg(x) = x - x^{2/3} - 2x^{2/3} + 2x^{1/3} + x^{1/3} - 1\]Further simplification:\[fg(x) = x - 3x^{2/3} + 3x^{1/3} - 1\]
07

Calculate the domain of (fg)(x)

As with \(f\) and \(g\), multiplication does not introduce any restrictions; the domain of \((fg)(x)\) remains all real numbers \((-\infty, \infty)\).
08

Calculate (f/g)(x)

To find \((f/g)(x)\), divide \(f(x)\) by \(g(x)\):\[(f/g)(x) = \frac{f(x)}{g(x)} = \frac{x^{2/3} - 2x^{1/3} + 1}{x^{1/3} - 1}\]
09

Calculate the domain of (f/g)(x)

For \((f/g)(x)\), the denominator \(x^{1/3} - 1\) must not be zero. This requires \(x eq 1\). Hence, the domain of \((f/g)(x)\) is all real numbers except \(x = 1\), or \( (-\infty, 1) \cup (1, \infty) \).

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Key Concepts

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

Function Addition
Adding two functions is like adding apples and oranges to get a fruit salad. When you add functions, you sum up their corresponding outputs for each input value. Given functions \( f(x) \) and \( g(x) \), the function addition is expressed as \((f+g)(x) = f(x) + g(x)\). For example, if \( f(x) = x^{2/3} - 2x^{1/3} + 1 \) and \( g(x) = x^{1/3} - 1 \), then:
  • Add the coefficients of like terms.
  • Combine all terms that have the same power.
By simplifying, we get:\[(f+g)(x) = x^{2/3} - x^{1/3}\]To determine the domain, remember that both \( f(x) \) and \( g(x) \) are defined for all real \( x \) because the expressions include cube roots, which are valid across all real numbers. Thus, the domain is \((-\infty, \infty)\).
Function Subtraction
Function Subtraction allows us to find the difference between two functions. It's similar to borrowing a cup of sugar from a neighbor. If you have \( f(x) \) and \( g(x) \), subtract \( g(x) \) from \( f(x) \) like so: \((f-g)(x) = f(x) - g(x)\). For our example functions:
  • Distribute the subtraction sign.
  • Subtract coefficients and simplify the expressions.
This gives us:\[(f-g)(x) = x^{2/3} - 3x^{1/3} + 2\]Similar to function addition, both original functions have no restriction over real numbers, which means the domain for \((f-g)(x)\) is \((-\infty, \infty)\).
Function Multiplication
Multiplying two functions is like planting seeds to grow new plants; you combine the values to create something new. For functions \( f(x) \) and \( g(x) \), you multiply the functions by applying the distributive property: \((fg)(x) = f(x) \cdot g(x)\). Consider our example functions:
  • Expand the product using distribution.
  • Simplify the expression by combining like terms.
This results in:\[(fg)(x) = x - 3x^{2/3} + 3x^{1/3} - 1\]Multiplication doesn鈥檛 restrict the domain further than the original functions, so the domain remains \((-\infty, \infty)\).
Function Division
Function Division helps us to divide one function by another, similar to slicing a pizza. For \( f(x) \) and \( g(x) \), division is expressed as \((f/g)(x) = \frac{f(x)}{g(x)}\). For given functions:
  • The key is to ensure the denominator isn't zero, as division by zero is undefined.
  • Identify where \( g(x) = 0 \) and exclude these from the domain.
So we have:\[(f/g)(x) = \frac{x^{2/3} - 2x^{1/3} + 1}{x^{1/3} - 1}\]To find the domain, set \( g(x) = x^{1/3} - 1 = 0 \), which means \( x = 1 \) must be excluded from the domain. Therefore, the domain is all real numbers except \( x = 1 \), or \((-\infty, 1) \cup (1, \infty)\).

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