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Magazine Chapter 3: Problem 68
The relationship between the dosage, \(x\), of a drug and the resulting change in body temperature is given by \(f(x)=x^{2}(3-x)\) for \(0 \leq x \leq 3 .\) Make sign diagrams for the first and second derivatives and sketch this dose- response curve, showing all relative extreme points and inflection points.
Short Answer
Expert verified
Max at \(x=0\); min at \(x=\frac{3}{2}\); inflection at \(x=\frac{3}{4}\).
Step by step solution
01
Determine the First Derivative
Calculate the first derivative of the function using the product and chain rules. The original function is given by \( f(x) = x^2(3-x) \). We apply the derivative rules as follows: first apply the product rule, and then simplify.\[ f'(x) = \frac{d}{dx}[x^2(3-x)] = 2x(3-x) + x^2(-1) = 6x - 3x^2 - x^2 = 6x - 4x^2. \]
02
Analyze the First Derivative
Set the first derivative equal to zero to find critical points. This will help us identify points where the slope changes, indicating potential maximums, minimums, or points of inflection.\[ 6x - 4x^2 = 0 \]\[ 2x(3-2x) = 0 \]\[ x = 0 \quad \text{or} \quad x = \frac{3}{2}. \] Construct a sign diagram to investigate these points and intervals \((0, \frac{3}{2}), (\frac{3}{2}, 3)\).
03
Determine the Second Derivative
Find the second derivative to analyze concavity and locate inflection points. Start by differentiating the first derivative.\[ f''(x) = \frac{d}{dx}(6x - 4x^2) = 6 - 8x. \]
04
Analyze the Second Derivative
Set the second derivative equal to zero to locate potential inflection points.\[ 6 - 8x = 0 \]\[ x = \frac{3}{4}. \] Construct a sign diagram to assess the concavity changes at \(x = \frac{3}{4}\) and intervals \((0, \frac{3}{4}), (\frac{3}{4}, 3)\).
05
Interpret the Sign Diagrams
From the first derivative sign diagram, \(x = 0\) changes from positive to negative, indicating a maximum, and \(x = \frac{3}{2}\) changes from negative to positive, indicating a minimum. The second derivative sign diagram confirms concave up to concave down at \(x \approx 0.75\), indicating an inflection point.
06
Sketch the Dose-response Curve
Using the insights from previous steps, sketch the curve with the maximum at \(x=0\), a minimum at \(x=\frac{3}{2}\), and an inflection point at \(x=\frac{3}{4}\). The function decreases initially, flattens at the maximum, then shows a change in concavity at the inflection point before reaching the minimum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
Understanding the first derivative is essential in calculus, especially when analyzing the behavior of functions like the drug dosage function provided in the exercise. The first derivative of a function, denoted as \( f'(x) \), provides the rate of change of the function's output with respect to its input. It tells us how the function is increasing or decreasing at a particular point.
In the given problem, we start with the function \( f(x) = x^2(3-x) \). Finding the first derivative involves using the product rule, which is used when differentiating a product of two functions, and we simplify using algebraic manipulations. This yields \( f'(x) = 6x - 4x^2 \).
The first derivative is crucial for identifying critical points, which occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. These points mark where the slope of the tangent to the curve is zero, indicating a peak (maximum), trough (minimum), or plateau. In this case, solving \( 6x - 4x^2 = 0 \) yields critical points at \( x = 0 \) and \( x = \frac{3}{2} \). This indicates potential locations for extreme values in the function.
In the given problem, we start with the function \( f(x) = x^2(3-x) \). Finding the first derivative involves using the product rule, which is used when differentiating a product of two functions, and we simplify using algebraic manipulations. This yields \( f'(x) = 6x - 4x^2 \).
The first derivative is crucial for identifying critical points, which occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. These points mark where the slope of the tangent to the curve is zero, indicating a peak (maximum), trough (minimum), or plateau. In this case, solving \( 6x - 4x^2 = 0 \) yields critical points at \( x = 0 \) and \( x = \frac{3}{2} \). This indicates potential locations for extreme values in the function.
Second Derivative
The second derivative of a function, written as \( f''(x) \), provides insight into the concavity of the function, which helps us understand how the slope of the tangent line is changing. To find it, we differentiate the first derivative.
For the given function, starting from \( f'(x) = 6x - 4x^2 \), we differentiate again to get \( f''(x) = 6 - 8x \). This second derivative is used to determine where the function is concave up (curved upwards) or concave down (curved downwards).
Setting \( f''(x) = 0 \) helps us find possible inflection points, where the concavity changes. Solving \( 6 - 8x = 0 \) gives us \( x = \frac{3}{4} \), indicating that concavity changes around this point. This information is crucial for accurately sketching the shape of the curve, showing how it bends and turns.
For the given function, starting from \( f'(x) = 6x - 4x^2 \), we differentiate again to get \( f''(x) = 6 - 8x \). This second derivative is used to determine where the function is concave up (curved upwards) or concave down (curved downwards).
Setting \( f''(x) = 0 \) helps us find possible inflection points, where the concavity changes. Solving \( 6 - 8x = 0 \) gives us \( x = \frac{3}{4} \), indicating that concavity changes around this point. This information is crucial for accurately sketching the shape of the curve, showing how it bends and turns.
Critical Points
Critical points are essential in the study of calculus because they help us find and describe relative maxima and minima of a function. These points occur where the first derivative is zero or undefined.
In our exercise, we determined the critical points by setting the first derivative \( 6x - 4x^2 \) to zero, resulting in \( x = 0 \) and \( x = \frac{3}{2} \).
- At \( x = 0 \), the function changes from increasing to decreasing, indicating a local maximum.- At \( x = \frac{3}{2} \), the function changes from decreasing to increasing, indicating a local minimum.
Understanding these points allows us to sketch the graph more accurately and predict where the function achieves its highest and lowest values within the given interval.
In our exercise, we determined the critical points by setting the first derivative \( 6x - 4x^2 \) to zero, resulting in \( x = 0 \) and \( x = \frac{3}{2} \).
- At \( x = 0 \), the function changes from increasing to decreasing, indicating a local maximum.- At \( x = \frac{3}{2} \), the function changes from decreasing to increasing, indicating a local minimum.
Understanding these points allows us to sketch the graph more accurately and predict where the function achieves its highest and lowest values within the given interval.
Inflection Points
Inflection points are where the function's concavity changes from being concave up to concave down or vice versa. They are crucial for understanding the overall shape of the graph.
In the exercise, the second derivative \( f''(x) = 6 - 8x \) is used to find where concavity changes. Setting \( 6 - 8x = 0 \), we find \( x = \frac{3}{4} \) to be an inflection point.
Analyzing intervals around this point, such as \( (0, \frac{3}{4}) \) and \( (\frac{3}{4}, 3) \), verifies that the function indeed shifts its concavity, confirming \( x = \frac{3}{4} \) as an inflection point.
This knowledge aids in sketching an accurate graph by showing where curves change direction, enhancing the understanding of the relationship between dosage and body temperature.
In the exercise, the second derivative \( f''(x) = 6 - 8x \) is used to find where concavity changes. Setting \( 6 - 8x = 0 \), we find \( x = \frac{3}{4} \) to be an inflection point.
Analyzing intervals around this point, such as \( (0, \frac{3}{4}) \) and \( (\frac{3}{4}, 3) \), verifies that the function indeed shifts its concavity, confirming \( x = \frac{3}{4} \) as an inflection point.
This knowledge aids in sketching an accurate graph by showing where curves change direction, enhancing the understanding of the relationship between dosage and body temperature.
