Chapter 3: Problem 46
Find \(\Delta y\) for the given values of \(x_{1}\) and \(x_{2}(\) see Example 7). $$ y=3 x^{2}+2 x+1, x_{1}=0.0, x_{2}=0.1 $$
Short Answer
Expert verified
\( \Delta y = 0.23 \)
Step by step solution
01
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 \( x_2 = 0.1 \).
02
Calculate \( y_{1} \) at \( x_{1} \)
Substitute \( x_1 = 0.0 \) into the function to find \( y_1 \): \[ y_1 = 3(0.0)^2 + 2(0.0) + 1 = 1 \]
03
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, \[ y_2 = 0.03 + 0.2 + 1 = 1.23 \]
04
Compute \( \Delta y \)
Subtract \( y_1 \) from \( y_2 \) to find \( \Delta y \): \[ \Delta y = y_2 - y_1 = 1.23 - 1 = 0.23 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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
When \( x = 0.1 \), by substituting \( x \) with \( 0.1 \), you'll determine \( y_2 \):
- \( y = 3(0.0)^2 + 2(0.0) + 1 \)
- Gives \( y = 1 \).
When \( x = 0.1 \), by substituting \( x \) with \( 0.1 \), you'll determine \( y_2 \):
- \( y_2 = 3(0.1)^2 + 2(0.1) + 1 = 1.23 \).
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:
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 \( x_2 = 0.1 \): \( y_2 = 1.23 \).
- \( \Delta y = y_2 - y_1 \)
- \( \Delta y = 1.23 - 1 = 0.23 \).
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:
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.