Problem 24 - Calculus Volume 1

Question

Set up a table to sketch the graph of each function using the following values: \(x=-3,-2,-1,0,1,2,3\) \(\quad f(x)=\frac{1}{2} x+1\)

Step-by-Step Solution

Answer

Sketch a table with \( x \) values and corresponding \( f(x) \) values.

Step 1: Identify the function

The given function is the linear function \( f(x) = \frac{1}{2}x + 1 \). This equation is in the slope-intercept form \( y = mx + b \), where \( m = \frac{1}{2} \) and \( b = 1 \).

Step 2: Create a table for x and f(x)

We'll evaluate the function \( f(x) \) for each specified \( x \) value: \(-3, -2, -1, 0, 1, 2, 3\). We'll fill in the table with these \( x \) values and their corresponding \( f(x) \) values.

Step 3: Calculate f(x) for each x value

- For \( x = -3 \): \( f(-3) = \frac{1}{2}(-3) + 1 = -\frac{3}{2} + 1 = -\frac{1}{2} \)- For \( x = -2 \): \( f(-2) = \frac{1}{2}(-2) + 1 = -1 + 1 = 0 \)- For \( x = -1 \): \( f(-1) = \frac{1}{2}(-1) + 1 = -\frac{1}{2} + 1 = \frac{1}{2} \)- For \( x = 0 \): \( f(0) = \frac{1}{2}(0) + 1 = 1 \)- For \( x = 1 \): \( f(1) = \frac{1}{2}(1) + 1 = \frac{1}{2} + 1 = \frac{3}{2} \)- For \( x = 2 \): \( f(2) = \frac{1}{2}(2) + 1 = 1 + 1 = 2 \)- For \( x = 3 \): \( f(3) = \frac{1}{2}(3) + 1 = \frac{3}{2} + 1 = \frac{5}{2} \)

Step 4: Record the values in the table

The table to sketch the graph is as follows:\[\begin{array}{|c|c|}\hlinex & f(x) \\hline-3 & -\frac{1}{2} \-2 & 0 \-1 & \frac{1}{2} \0 & 1 \1 & \frac{3}{2} \2 & 2 \3 & \frac{5}{2} \\hline\end{array}\]

Key Concepts

Slope-Intercept Form

The slope-intercept form is an important way to describe linear functions. It is expressed as \( y = mx + b \). Here, \( m \) is the slope, and \( b \) is the y-intercept. The slope of a line, \( m \), indicates how steep the line is.

• For example, a slope of \( \frac{1}{2} \) suggests that for every increase of 1 in \( x \), \( y \) increases by \( \frac{1}{2} \).
• Y-intercept, \( b \), is the point where the line crosses the y-axis. If \( b = 1 \), the line crosses the y-axis at \( y = 1 \).

These two parameters give us a complete picture to sketch the graph. Understanding slope-intercept form helps in easily identifying the behavior of a line by simply looking at its equation. This method simplifies graphing because it directly correlates with the linear graph's slope and starting point.

Graph Sketching

Graph sketching is a visual representation of the linear function. It involves plotting points calculated from the function and then drawing the line that connects these points.Begin by choosing specific \( x \) values. Here, they are \(-3, -2, -1, 0, 1, 2, 3\). Calculate \( f(x) \) for each of these x-values:

• For \( x = -3 \), \( f(-3) = -\frac{1}{2} \)
• For \( x = -2 \), \( f(-2) = 0 \)
• For \( x = -1 \), \( f(-1) = \frac{1}{2} \)
• For \( x = 0 \), \( f(0) = 1 \)
• For \( x = 1 \), \( f(1) = \frac{3}{2} \)
• For \( x = 2 \), \( f(2) = 2 \)
• For \( x = 3 \), \( f(3) = \frac{5}{2} \)

These pairs of \( (x, f(x)) \) are points on the graph.Once plotted, draw a straight line through these points, because the function \( f(x) = \frac{1}{2}x + 1 \) is linear. Graph sketching helps visualize linear relationships and how changes in \( x \) affect \( y \).

Evaluating Functions

Evaluating functions means finding the value of \( f(x) \) for specific \( x \) values. This process is crucial for understanding how input \( x \) relates to output \( f(x) \).Consider the function \( f(x) = \frac{1}{2}x + 1 \). To evaluate, substitute each x-value into the equation:

• First, multiply \( x \) by \( \frac{1}{2} \).
• Then, add 1 to the result.

Example: For \( x = 2 \), calculate \( f(2) = \frac{1}{2} \times 2 + 1 = 2 \).Repeat this for each specific x-value to find the corresponding \( f(x) \). This method provides a step-by-step approach to determine points for graphing or other analyses. Evaluating functions enables a deeper understanding of the function's behavior and how input directly influences the output.