Problem 30
Question
Let \(f(x)=2 x-5\) and \(g(x)=x+1 .\) Find each of the following function values. $$ (f-g)(-4) $$
Step-by-Step Solution
Verified Answer
The value of \((f-g)(-4)\) is \(-10\).
1Step 1: Understand the Function Difference
The difference \(f-g\)(x) is calculated by subtracting the function \(g(x)\) from \(f(x)\). This means \( (f-g)(x) = f(x) - g(x) \). Thus, we need to substitute the expressions for \(f(x)\) and \(g(x)\).
2Step 2: Substitute for f(x) and g(x)
We have \(f(x) = 2x - 5\) and \(g(x) = x + 1\). Substitute these into \(f(x) - g(x)\): \[(f-g)(x) = (2x - 5) - (x + 1)\]
3Step 3: Simplify the Expression
Simplify the expression from step 2:\[(f-g)(x) = 2x - 5 - x - 1\]Combine like terms:\[(f-g)(x) = (2x - x) + (-5 - 1) = x - 6\]
4Step 4: Evaluate (f-g)(-4)
Now substitute \(x = -4\) into the simplified expression \(x - 6\):\[(f-g)(-4) = -4 - 6\]Calculate the result:\[(f-g)(-4) = -10\]
Key Concepts
Understanding the Difference of FunctionsMastering Substitution in Function OperationsSimplification to Reach the Final Result
Understanding the Difference of Functions
The difference of functions is a fundamental concept in mathematics, especially when dealing with operations on functions. When you are asked to find the difference between two functions, denoted as \((f-g)(x)\), it means you need to subtract one function from the other. This involves taking the output of one function and subtracting the output of a second function.
For example, if you have two functions such as \(f(x) = 2x - 5\) and \(g(x) = x + 1\), finding their difference is as simple as performing the operation \(f(x) - g(x)\). It's crucial to express this operation clearly, as:
For example, if you have two functions such as \(f(x) = 2x - 5\) and \(g(x) = x + 1\), finding their difference is as simple as performing the operation \(f(x) - g(x)\). It's crucial to express this operation clearly, as:
- Identify each function: Know what \(f(x)\) and \(g(x)\) represent.
- Set up the difference: The expression becomes \( (f-g)(x) = f(x) - g(x) \).
Mastering Substitution in Function Operations
Substitution is a necessary skill for working with functions. It's the process of replacing the function expressions with their explicit formulas. Once you understand the difference of functions, as in \((f-g)(x) = f(x) - g(x)\), you are ready to substitute the actual expressions from the given functions.
Here’s how it works in our example with \(f(x) = 2x-5\) and \(g(x) = x+1\):
Here’s how it works in our example with \(f(x) = 2x-5\) and \(g(x) = x+1\):
- Take each function’s expression and place them into the set-up from the difference step.
- The equation becomes: \((f-g)(x) = (2x - 5) - (x + 1)\).
Simplification to Reach the Final Result
Simplification is the final step in solving function operations, and it involves breaking down the expression to its simplest form. After substitution, as seen in: \((f-g)(x) = (2x - 5) - (x + 1)\), you need to simplify these terms.
The steps are straightforward:
Finally, you evaluate this expression at the desired point, for instance when \(x = -4\). Substitute and calculate the result: \((f-g)(-4) = -4 - 6 = -10\). This touches all elements of function operations, making everything clear and concise!
The steps are straightforward:
- Eliminate parentheses by distributing negative signs across the subtracted terms.
- Combine like terms: \(2x - x\) simplifies to \(x\), and \(-5 - 1\) simplifies to \(-6\).
Finally, you evaluate this expression at the desired point, for instance when \(x = -4\). Substitute and calculate the result: \((f-g)(-4) = -4 - 6 = -10\). This touches all elements of function operations, making everything clear and concise!
Other exercises in this chapter
Problem 29
Evaluate expression. \(\log _{3} 3^{7}\)
View solution Problem 30
Solve each equation. Give the exact solution and an approximation to four decimal places. See Example 3. $$ 7^{x}=12 $$
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Graph each function. $$ y=e^{x-5} $$
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Write each logarithmic equation as an exponential equation. See Example 1. Do not solve. $$ \log _{8} \sqrt[3]{8}=\frac{1}{3} $$
View solution