Chapter 6: Problem 1
Find the limits of the function \(q=7-9 v+v^{2}:\) (a) As \(v \rightarrow 0\) (b) As \(v \rightarrow 3\) (c) As \(v \rightarrow-1\)
Short Answer
Expert verified
The limits are 7, -11, and 17 for \( v \to 0 \), \( v \to 3 \), and \( v \to -1 \) respectively.
Step by step solution
01
Understand the function
The given quadratic function is \( q = 7 - 9v + v^2 \). We need to find the limits as \( v \) approaches different values. This involves substituting the values into the function and simplifying.
02
Evaluate limit as v approaches 0
Substitute \( v = 0 \) into the function: \[q(0) = 7 - 9 \times 0 + 0^2 = 7.\]So, the limit \( \lim_{{v \to 0}} q(v) = 7 \).
03
Evaluate limit as v approaches 3
Substitute \( v = 3 \) into the function: \[q(3) = 7 - 9 \times 3 + 3^2 = 7 - 27 + 9 = -11.\]So, the limit \( \lim_{{v \to 3}} q(v) = -11 \).
04
Evaluate limit as v approaches -1
Substitute \( v = -1 \) into the function: \[q(-1) = 7 - 9 \times (-1) + (-1)^2 = 7 + 9 + 1 = 17.\]So, the limit \( \lim_{{v \to -1}} q(v) = 17 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
Limits are a fundamental concept in calculus that describe the behavior of a function as the input approaches a particular value. Essentially, limits tell us the value that a function approaches as the independent variable (in this case, \(v\)) gets closer and closer to a specific number. In our exercise, we're exploring how the quadratic function \(q = 7 - 9v + v^2\) behaves as \(v\) approaches 0, 3, and -1.
When evaluating limits, substituting the value directly into the function is a common method, provided the function is continuous at that point. In our example, for each of the specified values, substituting the number for \(v\) helps us determine the limit at those values as follows:
When evaluating limits, substituting the value directly into the function is a common method, provided the function is continuous at that point. In our example, for each of the specified values, substituting the number for \(v\) helps us determine the limit at those values as follows:
- For \(v \to 0\), substitute 0 into the function.
- For \(v \to 3\), substitute 3 into the function.
- For \(v \to -1\), substitute -1 into the function.
Quadratic Functions
Quadratic functions are polynomials of degree 2, typically written in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. Our function \(q = 7 - 9v + v^2\) is a quadratic function. It has the standard form \(v^2 - 9v + 7\), where \(a = 1\), \(b = -9\), and \(c = 7\).
Quadratic functions often form a parabolic curve on a graph, opening either upwards or downwards depending on the sign of the coefficient \(a\). In our example, since \(a = 1\) is positive, the parabola opens upwards. The vertex of a parabola is its highest or lowest point, and it can be a key focus point when analyzing the function's graph.
Quadratics are particularly interesting because they have different properties depending on their discriminant, \(b^2 - 4ac\):
Quadratic functions often form a parabolic curve on a graph, opening either upwards or downwards depending on the sign of the coefficient \(a\). In our example, since \(a = 1\) is positive, the parabola opens upwards. The vertex of a parabola is its highest or lowest point, and it can be a key focus point when analyzing the function's graph.
Quadratics are particularly interesting because they have different properties depending on their discriminant, \(b^2 - 4ac\):
- If it's positive, the quadratic has two distinct real roots.
- If it's zero, there is exactly one real root.
- If it's negative, the roots are complex and not visible on the real-number graph.
Substitution Method
The substitution method is an incredibly useful technique when assessing limits, especially when dealing with polynomials like quadratic functions. It involves replacing the variable in the function with the specified value, which helps determine the function's behavior as the variable approaches a particular number.
In our exercise, we use substitution directly:
In our exercise, we use substitution directly:
- For \(v \to 0\), substituting 0 gives us the limit \(7\).
- For \(v \to 3\), substituting 3 yields the limit \(-11\).
- For \(v \to -1\), substituting -1 results in the limit \(17\).