Problem 98 - Algebra for College Students

Question

Let \(f(x)=3 x-2, g(x)=-x^{2}+3 x-2,\) and \(h(x)=|x+2|\). Evaluate each expression. See Example 9. $$ f(1)-g(0) $$

Step-by-Step Solution

Answer

3

Step 1: Evaluate f(1)

The function given is \( f(x) = 3x - 2 \). To find \( f(1) \), substitute \( x = 1 \): \[ f(1) = 3(1) - 2 = 3 - 2 = 1. \]

Step 2: Evaluate g(0)

The function given is \( g(x) = -x^2 + 3x - 2 \). To find \( g(0) \), substitute \( x = 0 \): \[ g(0) = -(0)^2 + 3(0) - 2 = 0 + 0 - 2 = -2. \]

Step 3: Calculate f(1) - g(0)

Subtract the value of \( g(0) \) from \( f(1) \): \[ f(1) - g(0) = 1 - (-2) = 1 + 2 = 3. \]

Key Concepts

Evaluating Functions

Functions are like machines that take an input value and produce an output. We usually represent functions with letters like \(f\), \(g\), or \(h\). To evaluate a function, we simply replace the variable (often \(x\)) with the given number. For example, with \(f(x) = 3x - 2\) and you need to find \(f(1)\), you substitute \(x\) with 1.
Substituting values into a function helps us understand how the function behaves at specific points. It's a straightforward process but forms the foundation of more complex algebraic operations.

Substitution

In math, substitution is like swapping out one part of an equation for another. When evaluating functions, you're substituting the variable with a given number. For example, with \(g(x) = -x^2 + 3x - 2\), and you want to find \(g(0)\), you replace \(x\) with 0:
\((0)\) in place of \(x\) changes the equation to:

• \(-(0)^2 + 3(0) - 2 = 0 + 0 - 2 = -2\)

By substituting values into functions, you can evaluate and simplify expressions confidently.

Arithmetic Operations

After evaluating functions, arithmetic operations help combine results. Arithmetic includes addition, subtraction, multiplication, and division. For instance, once you've found \(f(1) = 1\) and \(g(0) = -2\), you subtract \(g(0)\) from \(f(1)\) for the final step:
That's done as follows:

• \(f(1) - g(0) = 1 - (-2) = 1 + 2 = 3\)

Combining these steps and operations allows us to solve complex equations easily. Practice makes these processes intuitive over time.