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Magazine Chapter 3: Problem 6
Find the relative extrema of the function, if they exist. ist your answers in terms of ordered pairs. Then sketch a graph of the function. $$ f(x)=0.5 x^{2}-2 x-11 $$
Short Answer
Expert verified
The function has a relative minimum at (2, -13).
Step by step solution
01
- Find the First Derivative
To find the relative extrema, first find the first derivative of the function. Given: The first derivative of the function can be found using the power rule: The derivative of: f(x) = 0.5x^2 鈭� 2x 鈭� 11 is f鈥�(x) = 0.5 * 2x 鈭� 2 = x 鈭� 2.
02
- Set the First Derivative Equal to Zero
To find critical points, set the first derivative equal to zero and solve for x. f鈥�(x) = x 鈭� 2 = 0. Solving for x gives: x = 2.
03
- Determine the Function Value at the Critical Point
To find the corresponding y-value, substitute x = 2 back into the original function. f(2) = 0.5 * (2)^2 鈭� 2 * 2 鈭� 11 = 2 鈭� 4 鈭� 11 = 鈭�13. So, the critical point is (2, -13).
04
- Determine the Nature of the Critical Point
To determine if the critical point is a minimum or maximum, check the second derivative. The second derivative of the function is: f''(x) = 1. Since f''(x) is positive, the critical point (2, -13) is a relative minimum.
05
- Sketch the Graph
To sketch the graph of the function: - Plot the critical point (2, -13), which is a relative minimum. - The function is a quadratic, and since the coefficient of x^2 is positive (0.5), the parabola opens upwards. - The vertex is at (2, -13), and the function is symmetric about the vertical line x = 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
The first derivative of a function helps us understand the rate at which the function's value is changing. When dealing with a quadratic function like \( f(x) = 0.5x^2 - 2x - 11 \), we can find the first derivative using basic differentiation rules.
For any term of the form \( ax^n \), its first derivative is \( anx^{(n-1)} \). Applying this to \( f(x) \):
For any term of the form \( ax^n \), its first derivative is \( anx^{(n-1)} \). Applying this to \( f(x) \):
- The derivative of \( 0.5x^2 \) is \( 0.5 \times 2x = x \)
- The derivative of \( -2x \) is \( -2 \)
- The derivative of a constant, \( -11 \), is \( 0 \)
Critical Points
Critical points are where the first derivative equals zero or it does not exist. These points are important because they can indicate relative maxima, minima, or points of inflection. To find these points for our function, set \( f'(x) = 0 \):
\( f(2) = 0.5 \times (2)^2 - 2 \times 2 - 11 = -13 \).
So, our critical point is \( (2, -13) \).
- Set the first derivative equal to zero: \( x - 2 = 0 \)
- Solve for \( x \): \( x = 2 \)
\( f(2) = 0.5 \times (2)^2 - 2 \times 2 - 11 = -13 \).
So, our critical point is \( (2, -13) \).
Second Derivative
The second derivative helps us understand the concavity of the function at the critical points. It can tell us whether a point is a relative maximum or minimum. For our function, consider the second derivative:
Since our first derivative is \( f'(x) = x - 2 \), differentiating this gives us:
\( f''(x) = 1 \).
The second derivative is constant and positive. When the second derivative is positive, the function is concave up at that point, indicating a relative minimum.
Therefore, the critical point \( (2, -13) \) is a relative minimum.
Since our first derivative is \( f'(x) = x - 2 \), differentiating this gives us:
\( f''(x) = 1 \).
The second derivative is constant and positive. When the second derivative is positive, the function is concave up at that point, indicating a relative minimum.
Therefore, the critical point \( (2, -13) \) is a relative minimum.
Quadratic Function
A quadratic function is of the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. For our function \( f(x) = 0.5x^2 - 2x - 11 \), quadratic functions graph as parabolas.
The coefficient of \( x^2 \) (in this case, 0.5) determines the shape and direction of the parabola:
The coefficient of \( x^2 \) (in this case, 0.5) determines the shape and direction of the parabola:
- If the coefficient is positive, the parabola opens upwards.
- If the coefficient is negative, the parabola opens downwards.
Graph Sketching
Sketching the graph of a quadratic function involves plotting the vertex and understanding the symmetry and direction of the parabola. Here's how to sketch \( f(x) = 0.5x^2 - 2x - 11 \):
- Plot the critical point, \( (2, -13) \), which is the vertex and a relative minimum.
- Since the coefficient of \( x^2 \) is positive, the parabola opens upwards.
- The function is symmetric around the vertical line \( x = 2 \).
