Problem 15
Question
Let \(f(x)=2 x+1\) and \(g(x)=x-3 .\) Find each function and give its domain. $$ g-f $$
Step-by-Step Solution
Verified Answer
The function is \(g-f = -x - 4\) with domain \(\mathbb{R}\).
1Step 1: Understand the Problem
We need to find the function that represents \(g-f\), which means we need to subtract function \(f(x)\) from \(g(x)\). We also need to determine the domain of this new function.
2Step 2: Substitute Functions
Substitute the given functions into the expression for \(g-f\). The expression becomes:\[g(x) - f(x) = (x-3) - (2x+1)\]
3Step 3: Simplify the Expression
Simplify the expression for \(g-f\):\[(x-3) - (2x+1) = x - 3 - 2x - 1 = -x - 4\]Thus, the function \(g-f\) is \(-x - 4\).
4Step 4: Determine the Domain
Both \(f(x) = 2x + 1\) and \(g(x) = x - 3\) are linear functions. Linear functions are defined for all real numbers. Therefore, the subtraction \(-x - 4\) is also defined for all real numbers. Thus, the domain of \(g-f\) is all real numbers, \(\mathbb{R}\).
Key Concepts
Subtracting FunctionsSimplifying ExpressionsDomain of Functions
Subtracting Functions
When dealing with functions in algebra, sometimes you have to perform operations such as addition, subtraction, multiplication, or division. Subtracting functions is a common operation and is essentially a straightforward process.
To subtract one function from another, say function \(f(x)\) from \(g(x)\), you simply take all of \(f(x)\) and subtract it from all of \(g(x)\). You need to pay attention to the signs while subtracting every term.
In our exercise, \(g(x) = x - 3\) and \(f(x) = 2x + 1\), the subtraction \(g-f\) involves subtracting each component of \(f(x)\) from the corresponding parts of \(g(x)\). This gives the expression:
To subtract one function from another, say function \(f(x)\) from \(g(x)\), you simply take all of \(f(x)\) and subtract it from all of \(g(x)\). You need to pay attention to the signs while subtracting every term.
In our exercise, \(g(x) = x - 3\) and \(f(x) = 2x + 1\), the subtraction \(g-f\) involves subtracting each component of \(f(x)\) from the corresponding parts of \(g(x)\). This gives the expression:
- \(g(x) - f(x) = (x - 3) - (2x + 1)\)
- Make sure to distribute the negative sign appropriately: \((x - 3) - 2x - 1\)
- Finally, simplify: \(-x - 4\).
Simplifying Expressions
Simplifying an expression means rewriting it in a simpler or more efficient form. This often involves combining like terms and following the order of operations.
For the subtraction of our functions, \( (x-3) - (2x+1) \), simplifying requires a few key steps:
For the subtraction of our functions, \( (x-3) - (2x+1) \), simplifying requires a few key steps:
- Remove parentheses. Just remember—be careful with signs when you distribute the subtraction: \(x - 3 - 2x - 1\).
- Next, combine like terms. Group all the \(x\)'s together and all constant terms together: \(x - 2x\) and \(-3 - 1\).
- This yields \(-x - 4\), which is the simplified version of our expression.
Domain of Functions
The domain of a function encompasses all the input values (\(x\) values) that the function can handle without encountering issues. For instance, you may encounter fractions where the denominator cannot be zero, or square roots must not take negative numbers.
However, both \(f(x) = 2x + 1\) and \(g(x) = x - 3\) are linear functions. A linear function is defined for every real number, meaning there are no restrictions like those seen in fractions or square roots.
Consequently, the subtraction \(g-f = -x - 4\) inherits this all-encompassing domain from its linear components.
However, both \(f(x) = 2x + 1\) and \(g(x) = x - 3\) are linear functions. A linear function is defined for every real number, meaning there are no restrictions like those seen in fractions or square roots.
Consequently, the subtraction \(g-f = -x - 4\) inherits this all-encompassing domain from its linear components.
- The domain for \(g-f\) is \(\mathbb{R}\), the set of all real numbers.
- This means the function never runs into any undefined areas, ensuring it can take on any real number as an input value without trouble.
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