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Magazine Chapter 3: Problem 75
For the given functions fand g. find the following. For parts \((a)-(d),\) also find the domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f \cdot g)(x)\) (d) \(\left(\frac{f}{g}\right)(x)\) (e) \((f+g)\) (3) (f) \((f-g)\) (4) (g) \((f \cdot g)\) ( 2 ) \((h)\left(\frac{f}{g}\right)(1)\) \(f(x)=\sqrt{x} ; g(x)=3 x-5\)
Short Answer
Expert verified
Domains: (a), (b), (c) -> \(x \ge 0\); (d) -> \(x \ge 0\) and \(x eq \frac{5}{3}\). Evaluations: (e) \(4 + \sqrt{3}\), (f) \(-5\), (g) \(\sqrt{2}\), (h) \(-\frac{1}{2}\).
Step by step solution
01
Understanding the Functions
Given functions are: \(f(x) = \sqrt{x} \) and \( g(x) = 3x - 5 \). We need to perform various operations on these functions and find their domains.
02
- Sum of Functions \((f+g)(x)\)
To find the sum of \(f\) and \(g\), add the functions: \[(f + g)(x) = f(x) + g(x) = \sqrt{x} + (3x - 5)\]The domain is the set of all x such that both \(\sqrt{x}\) and \(3x-5\) are defined. The domain for \(f(x) = \sqrt{x}\) is \(x \ge 0\). Hence, the domain for \((f+g)(x)\) is \(x \ge 0\).
03
- Difference of Functions \((f-g)(x)\)
To find the difference of \(f\) and \(g\), subtract the functions: \[(f - g)(x) = f(x) - g(x) = \sqrt{x} - (3x - 5) = \sqrt{x} - 3x + 5\]The domain is the same as in Step 1, as both functions \(\sqrt{x}\) and \(3x-5\) need to be defined. Therefore, the domain is \(x \ge 0\).
04
- Product of Functions \((f \cdot g)(x)\)
To find the product of \(f\) and \(g\), multiply the functions: \[(f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot (3x - 5) = 3x\sqrt{x} - 5\sqrt{x}\]The domain for this operation is still \(x \ge 0\).
05
- Quotient of Functions \((\frac{f}{g})(x)\)
To find the quotient of \(f\) and \(g\), divide the functions: \[\left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{3x - 5} \]The domain requires \(x \ge 0\) and also \(g(x) eq 0\). So, solve: \(3x - 5 eq 0\), which gives: \(x eq \frac{5}{3}\). Therefore, the domain is \(x \ge 0\) and \(x eq \frac{5}{3}\).
06
- Evaluate \((f+g)(3)\)
Find \((f + g)(3)\): \[(f + g)(3) = \sqrt{3} + (3 \cdot 3 - 5) = \sqrt{3} + 4 = \sqrt{3} + 4\]
07
- Evaluate \((f-g)(4)\)
Find \((f - g)(4)\): \[(f - g)(4) = \sqrt{4} - (3 \cdot 4 - 5) = 2 - 7 = -5\]
08
- Evaluate \((f \cdot g)(2)\)
Find \((f \cdot g)(2)\): \[(f \cdot g)(2) = 3 \cdot 2 \sqrt{2} - 5 \sqrt{2} = 6\sqrt{2} - 5\sqrt{2} = \sqrt{2}\]
09
- Evaluate \((\frac{f}{g})(1)\)
Find \( \left( \frac{f}{g} \right)(1) \): \[\left( \frac{f}{g} \right)(1) = \frac{\sqrt{1}}{3 \cdot 1 - 5} = \frac{1}{3 - 5} = \frac{1}{-2} = -\frac{1}{2}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sum of Functions
The sum of two functions, \(f(x)\) and \(g(x)\), is simply adding them together. If \(f(x) = \sqrt{x}\) and \(g(x) = 3x - 5\), then the sum function is:
\((f + g)(x) = \sqrt{x} + (3x - 5)\).
The domain of this function is determined by the domain of both individual functions. Since \(\sqrt{x}\) is defined when \(x \geq 0\), and \(3x - 5\) is defined everywhere, the overall domain for \((f + g)(x)\) is \(x \geq 0\).
This means you can only use non-negative values of \(x\) to ensure the function works properly.
\((f + g)(x) = \sqrt{x} + (3x - 5)\).
The domain of this function is determined by the domain of both individual functions. Since \(\sqrt{x}\) is defined when \(x \geq 0\), and \(3x - 5\) is defined everywhere, the overall domain for \((f + g)(x)\) is \(x \geq 0\).
This means you can only use non-negative values of \(x\) to ensure the function works properly.
Difference of Functions
The difference of two functions, \(f(x)\) and \(g(x)\), is obtained by subtracting \(g(x)\) from \(f(x)\). For \(f(x) = \sqrt{x}\) and \(g(x) = 3x - 5\), the difference function is:
\((f - g)(x) = \sqrt{x} - (3x - 5) = \sqrt{x} - 3x + 5\).
Similar to the sum function, both parts of this function must be valid. Hence, the domain for \((f - g)(x)\) remains \(x \geq 0\).
By respecting the domains of the individual functions, you ensure the difference function performs correctly across all valid inputs.
\((f - g)(x) = \sqrt{x} - (3x - 5) = \sqrt{x} - 3x + 5\).
Similar to the sum function, both parts of this function must be valid. Hence, the domain for \((f - g)(x)\) remains \(x \geq 0\).
By respecting the domains of the individual functions, you ensure the difference function performs correctly across all valid inputs.
Product of Functions
The product of two functions, \(f(x)\) and \(g(x)\), involves multiplying them together. Given \(f(x) = \sqrt{x}\) and \(g(x) = 3x - 5\), the product function is:
\((f \cdot g)(x) = \sqrt{x} \cdot (3x - 5) = 3x\sqrt{x} - 5\sqrt{x}\).
The domain requirement comes from \(\sqrt{x}\), which means \(x\) must be greater than or equal to zero. Therefore, the domain of \((f \cdot g)(x)\) is \(x \geq 0\).
Keep these domain constraints in mind to prevent undefined expressions during multiplication.
\((f \cdot g)(x) = \sqrt{x} \cdot (3x - 5) = 3x\sqrt{x} - 5\sqrt{x}\).
The domain requirement comes from \(\sqrt{x}\), which means \(x\) must be greater than or equal to zero. Therefore, the domain of \((f \cdot g)(x)\) is \(x \geq 0\).
Keep these domain constraints in mind to prevent undefined expressions during multiplication.
Quotient of Functions
The quotient of two functions, \(f(x)\) and \(g(x)\), involves dividing \(f(x)\) by \(g(x)\). With \(f(x) = \sqrt{x}\) and \(g(x) = 3x - 5\), the quotient function is:
\(\left( \frac{f}{g} \right)(x) = \frac{\sqrt{x}}{3x - 5}\).
The domain for this function is more restrictive because it needs both the numerator and the denominator to be defined, and the denominator cannot be zero. Thus, \(3x - 5 eq 0\) (leading to \(x eq \frac{5}{3}\)). Also, \(\sqrt{x}\) dictates that \(x \geq 0\). Therefore, the domain is \(x \geq 0\) and \(x eq \frac{5}{3}\).
Respect these constraints to avoid division by zero and ensure all parts of the function are valid.
\(\left( \frac{f}{g} \right)(x) = \frac{\sqrt{x}}{3x - 5}\).
The domain for this function is more restrictive because it needs both the numerator and the denominator to be defined, and the denominator cannot be zero. Thus, \(3x - 5 eq 0\) (leading to \(x eq \frac{5}{3}\)). Also, \(\sqrt{x}\) dictates that \(x \geq 0\). Therefore, the domain is \(x \geq 0\) and \(x eq \frac{5}{3}\).
Respect these constraints to avoid division by zero and ensure all parts of the function are valid.
Function Domain
The domain of a function is the set of all possible inputs (x-values) that the function can accept without causing any mathematical problems.
This could mean:
For example, in our given functions, \(f(x) = \sqrt{x}\) has a domain of \(x \geq 0\). This is because square roots of negative numbers are not defined in the set of real numbers.
Similarly, when combining functions, always identify the domain constraints from each separate function and apply them to the combined function.
This could mean:
- Ensuring no division by zero.
- Validating square roots do not get negative values.
- Checking that logarithms have positive arguments.
For example, in our given functions, \(f(x) = \sqrt{x}\) has a domain of \(x \geq 0\). This is because square roots of negative numbers are not defined in the set of real numbers.
Similarly, when combining functions, always identify the domain constraints from each separate function and apply them to the combined function.
