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Chapter 2: Problem 6

Find the relative extrema of each function, if they exist. List each extremum along with the \(x\) -value at which it occurs. Then sketch a graph of the function. $$F(x)=0.5 x^{2}+2 x-11$$

Short Answer

Expert verified
The relative minimum is at \((x, y) = (-2, -13)\).

Step by step solution

01

- Find the First Derivative

First, find the first derivative of the function. The first derivative will help identify critical points. Given:\[F(x) = 0.5x^2 + 2x - 11\]The first derivative is:\[F'(x) = d/dx (0.5x^2 + 2x - 11)\]\[F'(x) = 0.5 \times 2x + 2 = x + 2\]
02

- Find Critical Points

Set the first derivative equal to zero to find the critical points:\[F'(x) = x + 2 = 0\]Solve for \(x\):\[x = -2\]
03

- Determine the Nature of Critical Points

Use the second derivative test to determine if the critical point is a maximum, minimum, or neither. The second derivative is:\[F''(x) = d/dx (x + 2) = 1\]Since \(F''(x) > 0\), the critical point at \(x = -2\) is a relative minimum.
04

- Find the Function Value at the Critical Point

Substitute \(x = -2\) back into the original function to find the \(y\)-value:\[F(-2) = 0.5(-2)^2 + 2(-2) - 11\]\[F(-2) = 0.5(4) - 4 - 11\]\[F(-2) = 2 - 4 - 11 = -13\]So the relative minimum occurs at \((x, y) = (-2, -13)\).
05

- Sketch the Graph

To sketch the graph, plot the parabolic function \(F(x) = 0.5x^2 + 2x - 11\) and clearly mark the relative minimum at \( (-2, -13) \). Since the coefficient of \(x^2\) is positive, the parabola opens upwards.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points are key to understanding the behavior of a function. They occur where the first derivative of the function is zero or undefined. When the derivative is zero, it means the slope of the tangent line is flat at that point. In our example, we found the first derivative:
\[F'(x) = x + 2\]
Setting the first derivative equal to zero and solving for \(x\) gives us the critical point:
\[x + 2 = 0 \rightarrow x = -2\]
This indicates that \(x = -2\) is a critical point. Checking whether this point is a maximum, minimum, or neither requires further analysis, often involving additional tests like the second derivative test.
First Derivative
The first derivative of a function gives us the rate of change or slope of the function's graph. It is crucial for finding critical points and understanding the behavior of the function. In this example, the function is:
\[F(x) = 0.5x^2 + 2x - 11\]
Taking the first derivative, we get:
\[F'(x) = d/dx (0.5x^2 + 2x - 11) = x + 2\]
This tells us how the function changes at different points. Setting the first derivative to zero helps identify points where the function might have a relative extremum.
Second Derivative Test
The second derivative test helps determine the nature of a critical point (whether it's a maximum, minimum, or neither). We obtain the second derivative by differentiating the first derivative. For our function, the second derivative is:
\[F''(x) = d/dx(x + 2) = 1\]
In this test:
  • If \(F''(x) > 0\), the critical point is a relative minimum.
  • If \(F''(x) < 0\), the critical point is a relative maximum.
  • If \(F''(x) = 0\), the test is inconclusive.
In our example, since \(F''(x) = 1 > 0\), the critical point at \(x = -2\) is identified as a relative minimum.
This information helps us understand how the function behaves around that point.
Parabolic Graph
A parabolic graph represents a quadratic function and has a signature 'U' shape if the leading coefficient is positive and an 'n' shape if it's negative. For the function \(F(x) = 0.5x^2 + 2x - 11\), since the coefficient of \(x^2\) is positive (0.5), the parabola opens upwards.
Here's what you need to sketch it:
  • Identify the vertex: For our function, the vertex is found at the critical point \((-2, -13)\).
  • Determine the direction: The positive coefficient confirms the parabola opens upwards.
  • Plot additional points if necessary: This improves the accuracy of the graph.
With this understanding, you can sketch the graph, marking the relative minimum at \((-2, -13)\). This point is the lowest on the graph, making it the vertex of the upward-opening parabola.

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