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To prepare for Section \(9.5,\) review functions (Sections 3.8 and \(5.9)\) Find each of the following, given \(f(x)=80 x+2500\) and \(g(x)=150 x\) $$ (g-f)(10)[5.9] $$

Short Answer

Expert verified
\((g-f)(10) = -1800\)

Step by step solution

01

Understand the Functions

Given the functions:1. \( f(x) = 80x + 2500 \)2. \( g(x) = 150x \)These functions represent linear relationships between the variable \( x \) and their outputs.
02

Substitute x in the Functions

Calculate the value of each function at \( x = 10 \):1. \( f(10) = 80(10) + 2500 = 800 + 2500 = 3300 \)2. \( g(10) = 150(10) = 1500 \)
03

Subtract the Results

To find \((g-f)(10)\), subtract the result of \( f(10) \) from \( g(10) \):\( g(10) - f(10) = 1500 - 3300 = -1800 \)

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Key Concepts

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

Function Evaluation
Understanding how to evaluate a function is essential for solving problems involving functions. A function takes an input, processes it using a specific rule or formula, and gives an output. For instance, given the function formula \(f(x) = 80x + 2500\), when you evaluate \(f\) at \(x = 10\), you substitute \(10\) wherever you see \(x\). Here's how it works step by step:
  • Substitute 10: \(f(10) = 80(10) + 2500\)
  • Simplify the expression: \(80 \times 10 = 800\)
  • Add the constant: \(800 + 2500 = 3300\)
So, \(f(10) = 3300\). Understanding this process is crucial because it forms the basis for using functions in various scenarios.
Substitution
Substitution is a method used to replace a variable with a specific value. In mathematics, especially with functions, substitution allows us to find the output for a given input value.
  • Identify the variable to replace, in this case, \(x\).
  • Insert the desired input value into the function wherever the variable appears.
  • Simplify any arithmetic operations to find the result.
To illustrate, for \(g(x) = 150x\), by substituting \(x = 10\), we get:
  • Replace \(x\): \(g(10) = 150(10)\)
  • Perform the multiplication: \(150 \times 10 = 1500\)
Therefore, \(g(10) = 1500\). Substitution helps in evaluating functions, solving equations, and simplifying expressions.
Linear Relationships
Linear functions describe a straight-line relationship between the independent variable \(x\) and the dependent variable. The general form of a linear function is \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Linear functions have the following properties:
  • The graph is a straight line.
  • The rate of change (slope) is constant.
  • The relationship is described by a first-degree polynomial.
Let's consider the function \(f(x) = 80x + 2500\). Here, 80 is the slope, indicating how much the function value increases as \(x\) increases by 1. The number 2500 is the y-intercept, showing where the line crosses the y-axis. Such functions are ideal for modeling consistent, linear relationships.
By understanding linear relationships, you'll recognize patterns and predict outcomes effectively in real-world situations.

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