Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\Delta y\) for the given values of \(x_{1}\) and \(x_{2}(\) see Example 7). $$ y=3 x^{2}+2 x+1, x_{1}=0.0, x_{2}=0.1 $$

Step-by-Step Solution

Answer

\( \Delta y = 0.23 \)

Step 1: Understand the Function and Values

The function given is \( y = 3x^2 + 2x + 1 \). We need to find \( \Delta y \) by evaluating the change in \( y \) as \( x \) changes from \( x_1 = 0.0 \) to \( x_2 = 0.1 \).

Step 2: Calculate \( y_{1} \) at \( x_{1} \)

Substitute \( x_1 = 0.0 \) into the function to find \( y_1 \): \[ y_1 = 3(0.0)^2 + 2(0.0) + 1 = 1 \]

Step 3: Calculate \( y_{2} \) at \( x_{2} \)

Substitute \( x_2 = 0.1 \) into the function to find \( y_2 \): \[ y_2 = 3(0.1)^2 + 2(0.1) + 1 \] Calculate each part: - \( 3(0.1)^2 = 3(0.01) = 0.03 \) - \( 2(0.1) = 0.2 \) So, \[ y_2 = 0.03 + 0.2 + 1 = 1.23 \]

Step 4: Compute \( \Delta y \)

Subtract \( y_1 \) from \( y_2 \) to find \( \Delta y \): \[ \Delta y = y_2 - y_1 = 1.23 - 1 = 0.23 \]

Key Concepts

Function Evaluation

Function evaluation is a crucial concept in calculus and plays a pivotal role in understanding how functions behave with varying inputs. Consider a polynomial function, such as the one given in the exercise: \( y = 3x^2 + 2x + 1 \). To evaluate a function means determining its output, \( y \), by substituting a value for \( x \) into the equation. For instance, if you set \( x = 0.0 \), substituting this into the function gives

• \( y = 3(0.0)^2 + 2(0.0) + 1 \)
• Gives \( y = 1 \).

This process helps you find the specific output related to an input, like calculating \( y_1 \) and \( y_2 \) at specified \( x \) values.
When \( x = 0.1 \), by substituting \( x \) with \( 0.1 \), you'll determine \( y_2 \):

• \( y_2 = 3(0.1)^2 + 2(0.1) + 1 = 1.23 \).

Evaluating functions provides key insights into how a graph behaves at specific points.

Change in y-value

Change in \( y \)-value, denoted as \( \Delta y \), refers to the difference in the output values of a function at two distinct input values. In the context of this exercise, finding \( \Delta y \) involves calculating the outputs of the function for \( x_1 \) and \( x_2 \) and measuring the difference. This change reflects how much the function's output varies as the input shifts from \( x_1 \) to \( x_2 \).

Start by calculating \( y_1 \) and \( y_2 \) for \( x_1 = 0.0 \) and \( x_2 = 0.1 \), respectively:

• For \( x_1 = 0.0 \): \( y_1 = 1 \).
• For \( x_2 = 0.1 \): \( y_2 = 1.23 \).

Then calculate \( \Delta y \) as follows:

• \( \Delta y = y_2 - y_1 \)
• \( \Delta y = 1.23 - 1 = 0.23 \).

This calculation helps you understand the behavior of the function as the input changes, showing the rate or magnitude of change in the output.

Polynomial Functions

Polynomial functions are mathematical expressions involving powers of \( x \), and they have coefficients and constant terms. The general form for a polynomial function is \( p(x) = a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 \), where \( a_n, a_{n-1}, ..., a_0 \) are constants and \( n \) is a non-negative integer.

In our example, the function \( y = 3x^2 + 2x + 1 \) is a second-degree polynomial because the highest exponent of \( x \) is 2. Each part of the polynomial contributes uniquely:

• The term \( 3x^2 \) describes a parabola opening upwards, scaled by 3.
• The term \( 2x \) influences the slope or steepness of the graph.
• The constant \( 1 \) vertically shifts the entire graph upwards by 1 unit.

Understanding polynomial functions is essential because they serve as a foundation for more complex mathematical concepts. They model many real-world scenarios and are especially useful in calculus for analysis and prediction.