Problem 29
Question
In Exercises \(29-36 :\)
a. Identify the function's local extreme values in the given domain, and say
where they are assumed.
b. Which of the extreme values, if any, are absolute?
c. Support your findings with a graphing calculator or computer grapher.
$$
f(x)=2 x-x^{2}, \quad-\infty
Step-by-Step Solution
Verified Answer
Local maximum at \(x = 1\) and absolute maximum at \(x = 1\); absolute minimum at \(x = 2\).
1Step 1: Find Critical Points
To find the critical points, we start by calculating the derivative of the function. The function given is \(f(x) = 2x - x^2\). The derivative is \(f'(x) = 2 - 2x\). We set \(f'(x) = 0\) to identify possible local extrema. Hence, \(2 - 2x = 0\) implies \(x = 1\).
2Step 2: Determine Local Extrema
To determine whether \(x = 1\) is a local minimum or maximum, we use the second derivative test. The second derivative of \(f(x)\) is \(f''(x) = -2\), which is negative, indicating that the function is concave down at \(x = 1\), suggesting a local maximum at this point.
3Step 3: Evaluate the Endpoints
Since the domain is \(-\infty < x \leq 2\), we must consider the endpoint at \(x = 2\). Evaluate the function at this point: \(f(2) = 2(2) - (2)^2 = 4 - 4 = 0\). At the critical point, \(f(1) = 2(1) - (1)^2 = 2 - 1 = 1\).
4Step 4: Compare to Find Absolute Extrema
Compare the function values at the local maximum and the endpoint. We have \(f(1) = 1\) and \(f(2) = 0\). Therefore, the highest value is \(1\) at \(x = 1\), and this is the absolute maximum within the given domain. The endpoint \(x = 2\) is the absolute minimum as \(f(x)\) decreases towards negative infinity beyond \(x = 1\).
5Step 5: Graphical Verification
Using a graphing calculator or computer software, plot \(f(x) = 2x - x^2\) within the domain \(-\infty < x \leq 2\). Verify that the graph reaches its peak at \(x = 1\) and then decreases to \(x = 2\). This visual confirmation supports our analytical findings.
Key Concepts
Critical PointsExtremaSecond Derivative TestConcavity
Critical Points
Critical points are vital landmarks on a function's curve where the slopes of tangent lines are zero or undefined. To find these points, start by determining the derivative of the function. Set this derivative equal to zero to establish where these slopes vanish. For example, given a function like \( f(x) = 2x - x^2 \), calculate its derivative \( f'(x) = 2 - 2x \). Setting \( f'(x) = 0 \), we solve \( 2 - 2x = 0 \), which gives us \( x = 1 \).
This means the critical point is at \( x = 1 \). Such points are not just about local hills and valleys; they can help reveal a function's behavior over its domain. Understanding critical points is a key step for identifying extreme values.
This means the critical point is at \( x = 1 \). Such points are not just about local hills and valleys; they can help reveal a function's behavior over its domain. Understanding critical points is a key step for identifying extreme values.
Extrema
Once critical points are found, we explore if they correspond to local or absolute extrema, i.e., the peaks and valleys of the function graph. Extrema refer to maximum or minimum values, either locally (in a small region) or absolutely (throughout the entire domain).
Local maxima or minima can occur at critical points or within the boundaries of the domain. For \( f(x) = 2x - x^2 \) in the domain \(-\infty
Local maxima or minima can occur at critical points or within the boundaries of the domain. For \( f(x) = 2x - x^2 \) in the domain \(-\infty
Second Derivative Test
To accurately classify whether a critical point is a local maximum, minimum, or neither, the second derivative test comes in handy. This test evaluates the curvature of the function by using its second derivative. If the second derivative at a critical point is positive, the function has a local minimum there. If it’s negative, there is a local maximum, indicating the slope transitions from positive to negative.
For the function \( f(x) = 2x - x^2 \), the second derivative is \( f''(x) = -2 \). Since \( f''(x) = -2 \) is negative at \( x = 1 \), the point is a local maximum. This verification aids in making precise assessments of critical points' nature.
For the function \( f(x) = 2x - x^2 \), the second derivative is \( f''(x) = -2 \). Since \( f''(x) = -2 \) is negative at \( x = 1 \), the point is a local maximum. This verification aids in making precise assessments of critical points' nature.
Concavity
Concavity describes the direction the graph of a function opens, either upward (concave up) like a bowl or downward (concave down) like a dome. The sign of the second derivative of a function can tell us about concavity. If \( f''(x) > 0 \), the graph is concave up, resembling a cup and usually indicating a local minimum. Conversely, if \( f''(x) < 0 \), the graph is concave down, indicating a local maximum within that region.
For the quadratic function \( f(x) = 2x - x^2 \), with \( f''(x) = -2 \), the graph is concave down throughout. Thus, any critical point found within this concavity will be a maximum, as was seen at \( x = 1 \) in the function's analysis. Understanding concavity helps in predicting and verifying the behavior and trend of the curve.
For the quadratic function \( f(x) = 2x - x^2 \), with \( f''(x) = -2 \), the graph is concave down throughout. Thus, any critical point found within this concavity will be a maximum, as was seen at \( x = 1 \) in the function's analysis. Understanding concavity helps in predicting and verifying the behavior and trend of the curve.
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