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The first derivative \(f^{\prime}\) is given. Find all values of \(x\) that make the function \(f(a)\) a local minimum and \((b)\) a local maximum. \(f^{\prime}(x)=-(x-1)(x-2)(x-3)(x-4)\)

Short Answer

Expert verified
Local minima at \(x = 2\) and \(4\); local maxima at \(x = 1\) and \(3\).

Step by step solution

01

Determine Critical Points

To find the critical points, set the first derivative equal to zero: \[-(x-1)(x-2)(x-3)(x-4) = 0\]. This gives us the potential critical points as solutions to \((x-1) = 0\), \((x-2) = 0\), \((x-3) = 0\), and \((x-4) = 0\). This leads to the critical points: \(x = 1, 2, 3, 4\).
02

Analyze Sign Changes in Derivative

Investigate how the sign of \(f^{\prime}(x)\) changes around each critical point to determine if it corresponds to a local maximum or minimum. Pick test points in each interval divided by these critical points, e.g., \(-\infty < x < 1\), \(1 < x < 2\), etc., to evaluate the sign transitions.
03

Sign Analysis in Intervals

Evaluate the sign of each factor between the critical points:- For \(x < 1\): All factors \((x-1), (x-2), (x-3), (x-4) < 0\), so \(f^{\prime}(x) > 0\).- For \(1 < x < 2\): \((x-1) > 0\), the others are negative, so \(f^{\prime}(x) < 0\).- For \(2 < x < 3\): \((x-1) > 0, (x-2) > 0\), the rest negative, so \(f^{\prime}(x) > 0\).- For \(3 < x < 4\): \((x-1), (x-2), (x-3) > 0\), \((x-4) < 0\), so \(f^{\prime}(x) < 0\).- For \(x > 4\): All factors positive, hence \(f^{\prime}(x) > 0\).
04

Identify Local Minima and Maxima

From the sign analysis:- At \(x = 1\), \(f^{\prime}(x) ext{ changes from positive to negative}\), indicating a local maximum.- At \(x = 2\), \(f^{\prime}(x) ext{ changes from negative to positive}\), indicating a local minimum.- At \(x = 3\), \(f^{\prime}(x) ext{ changes from positive to negative}\), indicating a local maximum.- At \(x = 4\), \(f^{\prime}(x) ext{ changes from negative to positive}\), indicating a local minimum.

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Key Concepts

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

Local Minima and Maxima
In calculus, local minima and maxima are essential concepts when studying the behavior of functions. These points help us determine where a function attains its highest or lowest values within a certain interval. A local minimum occurs when a function's value is smaller than all nearby points, while a local maximum is when a function's value is larger than all nearby points.

Identifying these points can help us understand and interpret the overall shape of the graph. For instance, in economics or biology, local extremum points might represent optimal conditions or critical thresholds. Understanding local minima and maxima allows us to prepare and anticipate changes in the function's behavior that affect real-world applications.
First Derivative Test
The first derivative test is a vital tool in calculus that helps identify local minima and maxima. It involves determining the critical points of a function by setting its first derivative to zero. These critical points are potential candidates for local minima and maxima. Once these points are identified, we can examine how the derivative behaves around these points.

By observing the sign of the derivative before and after a critical point, we can determine the nature of that critical point. For example:
  • If the derivative changes from positive to negative, it indicates a local maximum.
  • If the derivative changes from negative to positive, it implies a local minimum.
This test allows us to sort out these important changes without examining the entire function, making it extremely useful for quick analysis.
Sign Analysis
Sign analysis involves studying the sign of the first derivative over various intervals to identify how a function behaves in those segments. This process helps us confirm whether the critical points are indeed minima or maxima. During sign analysis, we divide the real number line into intervals marked by critical points.

For each interval, a test point is selected. By determining the sign of the first derivative at that test point, we can predict the function's behavior in that entire interval:
  • A positive derivative indicates the function is increasing.
  • A negative derivative means the function is decreasing.
This systematic approach allows for a clear understanding of the function's slope and helps identify local extrema effectively.
Calculus Problems
Calculus problems often involve finding aspects like local minima and maxima. They are not just about applying rules but understanding the changes and behaviors of functions. Using the concepts like the first derivative test and sign analysis, students solve practical problems, from optimizing engineering designs to predicting the spread of diseases.

These problems build a student's capacity to think critically and analytically. By breaking down complex issues into smaller parts, such as derivatives, limits, and integrals, calculus problems offer an insightful view into continuous change. They train students to see patterns, make predictions, and solve real-life situations by comprehensively understanding the underlying mathematics.

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\mathrm{\\{} E X P L ~ B e s i d e s ~ p r o v i d i n g ~ a n ~ e a s y ~ w a y ~ t o ~ d i f f e r e n t i a t e ~ p r o d u c t s , ~ l o g . ~ arithmic differentiation also provides a measure of the relative or fractional rate of change, defined as \(y^{\prime} / y .\) We explore this concept in Problems 39-42. Show that the relative rate of change of \(e^{k t}\) as a function of \(t\) is \(k\)

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