Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 45-50, find \(\Delta y\) for the given values of \(x_{1}\) and \(x_{2}\) (see Example 7). 45\. \(y=3 x+2, x_{1}=1, x_{2}=1.5\)

Step-by-Step Solution

Answer

\(\Delta y = 1.5\)

Step 1: Define the function

The given function is linear: \(y = 3x + 2\). This defines how \(y\) changes with \(x\).

Step 2: Calculate \(y\) at \(x_1\)

Substitute \(x_1 = 1\) into the function to find the value of \(y\). \[ y_1 = 3(1) + 2 = 3 + 2 = 5 \]

Step 3: Calculate \(y\) at \(x_2\)

Substitute \(x_2 = 1.5\) into the function to find the value of \(y\). \[ y_2 = 3(1.5) + 2 = 4.5 + 2 = 6.5 \]

Step 4: Compute \(\Delta y\)

Find the change in \(y\) by subtracting \(y_1\) from \(y_2\). \[ \Delta y = y_2 - y_1 = 6.5 - 5 = 1.5 \]

Key Concepts

Change in y

The concept of "change in y," often expressed as \( \Delta y \), represents how much the value of \( y \) changes as \( x \) shifts from one point to another. In linear functions, like the one we are dealing with here \( y = 3x + 2 \), this change is linear, meaning it varies predictably with changes in \( x \).
To determine \( \Delta y \), you calculate the difference between the values of \( y \) at two different \( x \) values. For instance, in our problem, we found \( y \) at \( x_1 = 1 \) to be 5, and at \( x_2 = 1.5 \) to be 6.5. Subsequently, the change in \( y \) is calculated by subtracting the initial value from the final value: \[ \Delta y = y_2 - y_1 = 6.5 - 5 = 1.5 \] This tells us that as \( x \) increased from 1 to 1.5, the corresponding \( y \) increased by 1.5.

Substitution Method

The substitution method is a crucial technique for solving both linear equations and understanding their behavior. It involves replacing a variable with a given value to find another quantity.

• In our problem, the function \( y = 3x + 2 \) requires substituting the \( x \) values to evaluate \( y \).
• First, for \( x_1 = 1 \), plugging this into the equation gives: \( y_1 = 3(1) + 2 = 5 \).
• Next, substituting \( x_2 = 1.5 \) results in: \( y_2 = 3(1.5) + 2 = 6.5 \).

Through substitution, the actual outputs of \( y \) at specific points are determined, allowing us to trace the function's progression as \( x \) changes.

Function Evaluation

Function evaluation is the process of calculating the output of a function for a specific input value. It plays a key role in understanding linear functions, as seen with the equation \( y = 3x + 2 \). By evaluating this function:

• We identify the specific output \( y \) at each \( x \) value.
• For \( x_1 = 1 \), subbing in gives \( y_1 = 5 \).
• For \( x_2 = 1.5 \), subbing provides \( y_2 = 6.5 \).

Evaluating the function at these points not only informs us of specific \( y \)-values but highlights how smoothly \( y \) transitions in response to changes in \( x \). It underlines the predictability and linearity of such functions.

Difference Calculation

Difference calculation is pivotal for understanding changes between two points, especially in linear functions. This involves subtracting one value from another to reveal the magnitude of change.

• Here, to find \( \Delta y \), we calculated the difference between \( y_2 \) and \( y_1 \): \( 6.5 - 5 = 1.5 \).
• This difference tells us the rise or increase in \( y \) as \( x \) shifts from 1 to 1.5.

Difference calculation helps not only in determining how variables shift but also in interpreting their behavior over specific intervals, ensuring a clear grasp of linear relationships.