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Magazine Chapter 2: Problem 3
Each of the graphs of the functions has one relative maximum and one relative minimum point. Find these points using the first-derivative test. Use a variation chart as in Example 1. $$f(x)=-x^{3}+6 x^{2}-9 x+1$$
Short Answer
Expert verified
Relative minimum: (1, -3). Relative maximum: (3, 1).
Step by step solution
01
- Find the first derivative
The first step is to find the first derivative of the function. The function given is:\[ f(x) = -x^3 + 6x^2 - 9x + 1 \]To find the first derivative, differentiate each term with respect to x:\[ f'(x) = -3x^2 + 12x - 9 \]
02
- Find the critical points
Set the first derivative equal to zero to find the critical points:\[ -3x^2 + 12x - 9 = 0 \]Solve this quadratic equation. Dividing through by -3 gives:\[ x^2 - 4x + 3 = 0 \]Factorize the quadratic equation:\[ (x - 3)(x - 1) = 0 \]The solutions are: \[ x = 3 \] and \[ x = 1 \]
03
- Create a variation chart
To determine whether these critical points are relative maxima or minima, build a variation chart. Evaluate the sign of the first derivative in the intervals determined by the critical points \[ x=1 \] and \[ x=3 \]:- Choose a test point in the interval \((-\infty, 1)\), for example, \( x = 0 \). Then, \[ f'(0) = -3(0)^2 + 12(0) - 9 = -9 \], which is negative.- Choose a test point in the interval \((1, 3)\), for example, \( x = 2 \). Then, \[ f'(2) = -3(2)^2 + 12(2) - 9 = -12 + 24 - 9 = 3 \], which is positive.- Choose a test point in the interval \((3, \infty)\), for example, \( x = 4 \). Then, \[ f'(4) = -3(4)^2 + 12(4) - 9 = -48 + 48 - 9 = -9 \], which is negative.
04
- Determine the nature of the critical points
Based on the variation chart, we can determine the nature of the critical points:- At \( x = 1 \), the derivative changes from negative to positive, indicating a relative minimum.- At \( x = 3 \), the derivative changes from positive to negative, indicating a relative maximum.
05
- Find the coordinates of the relative extrema
Substitute the critical points back into the original function to find the coordinates of the relative extrema:- For the relative minimum at \( x = 1 \):\[ f(1) = -(1)^3 + 6(1)^2 - 9(1) + 1 = -1 + 6 - 9 + 1 = -3 \]- For the relative maximum at \( x = 3 \):\[ f(3) = -(3)^3 + 6(3)^2 - 9(3) + 1 = -27 + 54 - 27 + 1 = 1 \]So, the coordinates are (1, -3) for the minimum and (3, 1) for the maximum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
relative extrema
Relative extrema refer to the relative maximum and minimum points of a function on a certain interval.
These points are where a function changes direction from increasing to decreasing or vice versa.
To identify relative extrema, we often use the first-derivative test.
This method involves finding the critical points of a function and then determining whether each critical point is a maximum, minimum, or neither.
These points are where a function changes direction from increasing to decreasing or vice versa.
To identify relative extrema, we often use the first-derivative test.
This method involves finding the critical points of a function and then determining whether each critical point is a maximum, minimum, or neither.
critical points
Critical points are values of x where the first derivative of a function is zero or undefined.
These points are important because they might be where the function has relative extrema.
To find the critical points, follow these steps:
These points are important because they might be where the function has relative extrema.
To find the critical points, follow these steps:
- Take the first derivative of the function.
- Set the first derivative equal to zero.
- Solve the resulting equation to find the x-values.
variation chart
A variation chart helps us determine the nature of the critical points (whether they are maxima, minima, or neither).
To create a variation chart, we do the following:
For \(x = 0\) in \((-\to 1)\), \(f'(0) = -9\) (negative); for \(x = 2\) in \((1, 3)\), \(f'(2) = 3\) (positive); for \(x = 4\) in \((3, \to)\), \(f'(4) = -9\) (negative). The sign changes indicate the relative extrema: a minimum at \(x = 1\) and a maximum at \(x = 3\).
To create a variation chart, we do the following:
- Determine the sign of the derivative in the intervals divided by the critical points.
- Choose test points within each interval and substitute them into the first derivative to find whether the derivative is positive or negative.
For \(x = 0\) in \((-\to 1)\), \(f'(0) = -9\) (negative); for \(x = 2\) in \((1, 3)\), \(f'(2) = 3\) (positive); for \(x = 4\) in \((3, \to)\), \(f'(4) = -9\) (negative). The sign changes indicate the relative extrema: a minimum at \(x = 1\) and a maximum at \(x = 3\).
derivative
The derivative of a function measures the rate at which the function's value changes as its input changes.
The first derivative, denoted as \(f'(x)\) or \(\frac{df}{dx}\), helps identify where the function increases or decreases.
Calculating the first derivative involves differentiating each term of the function. For instance, for the function \(f(x) = -x^3 + 6x^2 - 9x + 1\), the first derivative is \(f'(x) = -3x^2 + 12x - 9\).
We then use this derivative to find the critical points by setting \(f'(x) = 0\) and solving for x.
This step is crucial in analyzing the function's behavior and finding its relative extrema.
The first derivative, denoted as \(f'(x)\) or \(\frac{df}{dx}\), helps identify where the function increases or decreases.
Calculating the first derivative involves differentiating each term of the function. For instance, for the function \(f(x) = -x^3 + 6x^2 - 9x + 1\), the first derivative is \(f'(x) = -3x^2 + 12x - 9\).
We then use this derivative to find the critical points by setting \(f'(x) = 0\) and solving for x.
This step is crucial in analyzing the function's behavior and finding its relative extrema.
