Problem 39 - Intermediate Algebra

Question

If \(f(x)=\frac{1}{2} x-\frac{3}{4}\), find \(f(-2), f(0), f\left(\frac{1}{2}\right), f\left(\frac{2}{3}\right)\)

Step-by-Step Solution

Answer

\(f(-2) = -\frac{7}{4}, f(0) = -\frac{3}{4}, f\left(\frac{1}{2}\right) = -\frac{1}{2}, f\left(\frac{2}{3}\right) = -\frac{5}{12}\).

Step 1: Understand the Function

The function given is a linear function: \( f(x) = \frac{1}{2}x - \frac{3}{4} \). It's a line equation where \( \frac{1}{2} \) is the slope and \( -\frac{3}{4} \) is the y-intercept.

Step 2: Substitute \( x = -2 \)

To find \( f(-2) \), substitute \( -2 \) for \( x \) in the function: \( f(-2) = \frac{1}{2}(-2) - \frac{3}{4} = -1 - \frac{3}{4} = -\frac{7}{4} \).

Step 3: Substitute \( x = 0 \)

To find \( f(0) \), substitute \( 0 \) for \( x \) in the function: \( f(0) = \frac{1}{2}(0) - \frac{3}{4} = -\frac{3}{4} \).

Step 4: Substitute \( x = \frac{1}{2} \)

To find \( f\left(\frac{1}{2}\right) \), substitute \( \frac{1}{2} \) for \( x \) in the function: \( f\left(\frac{1}{2}\right) = \frac{1}{2}\left(\frac{1}{2}\right) - \frac{3}{4} = \frac{1}{4} - \frac{3}{4} = -\frac{1}{2} \).

Step 5: Substitute \( x = \frac{2}{3} \)

To find \( f\left(\frac{2}{3}\right) \), substitute \( \frac{2}{3} \) for \( x \) in the function: \( f\left(\frac{2}{3}\right) = \frac{1}{2}\left(\frac{2}{3}\right) - \frac{3}{4} = \frac{1}{3} - \frac{3}{4} = -\frac{5}{12} \).

Key Concepts

Function Evaluation

In the world of mathematics, functions are like machines where you put in numbers and get results based on the rule given by the function. "Function evaluation" is the process of finding the output of that function when a specific input is given.
To evaluate a function, you simply substitute the input value (also known as the "argument") into the function. These inputs are often represented by the variable \( x \). For example, if you want to find \( f(-2) \) for the function \( f(x) = \frac{1}{2}x - \frac{3}{4} \), you replace \( x \) with \( -2 \) and compute the result.

• Start by identifying the function rule.
• Replace the variable \( x \) with the given input.
• Perform the arithmetic to find the final value.

Understanding function evaluation is crucial because it helps you see how changes in \( x \) affect the output, or \( y \), in the function.

Substitution Method

The substitution method is a straightforward technique used to find the value of a function for a given input. In essence, it involves replacing variables with given numbers or expressions.
When using the substitution method in our function \( f(x) = \frac{1}{2}x - \frac{3}{4} \), you plug in specific values for \( x \). For example:

• To find \( f(0) \), substitute \( 0 \) for \( x \), resulting in \( f(0) = \frac{1}{2}(0) - \frac{3}{4} = -\frac{3}{4} \).
• To find \( f\left(\frac{1}{2}\right) \), substitute \( \frac{1}{2} \) for \( x \), giving \( f\left(\frac{1}{2}\right) = \frac{1}{2}\cdot\frac{1}{2} - \frac{3}{4} = -\frac{1}{2} \).

This method can be used for solving equations or checking answers by replacing variables with specific numbers and simplifying the equation. It's a handy process whenever you have an expression containing variables.

Slope-Intercept Form

The slope-intercept form is a way to write the equation of a straight line. It is generally expressed as \( y = mx + b \), where \( m \) represents the slope of the line, and \( b \) represents the y-intercept.
In our example, the function \( f(x) = \frac{1}{2}x - \frac{3}{4} \) is already in slope-intercept form.

• The slope (\( m \)) is \( \frac{1}{2} \). It shows how much \( y \) changes for a one-unit change in \( x \). A positive slope means the line rises as \( x \) increases, while a negative slope means it falls.
• The y-intercept (\( b \)) is \( -\frac{3}{4} \). This is the point where the line crosses the y-axis. In simpler terms, it's the value of \( f(x) \) when \( x \) is 0.

By understanding the slope and y-intercept, you can quickly grasp how the line behaves without having to plot every point. This is especially useful for graphing and analyzing linear relationships.