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Chapter 4: Problem 13

Find all critical points and then use the second-derivative test to determine local maxima and minima. $$f(x)=9+6 x^{2}-x^{3}$$

Short Answer

Expert verified
x = 0 is a local minimum; x = 4 is a local maximum.

Step by step solution

01

Find the First Derivative

To find the critical points, we first need to compute the first derivative of the function.The given function is:\[ f(x) = 9 + 6x^2 - x^3 \]Using differentiation, we find the first derivative:\[ f'(x) = \frac{d}{dx}(9) + \frac{d}{dx}(6x^2) - \frac{d}{dx}(x^3) \]\[ f'(x) = 0 + 12x - 3x^2 \]Thus, the first derivative is:\[ f'(x) = 12x - 3x^2 \].
02

Solve for Critical Points

Critical points occur where the first derivative is zero or undefined. We set the first derivative to zero and solve for \( x \):\[ 12x - 3x^2 = 0 \]Factor out the common term:\[ 3x(4 - x) = 0 \]This gives us the potential critical points:\[ x = 0 \]and\[ x = 4 \].
03

Second Derivative Test

Next, we use the second derivative to determine the nature of the critical points. First, we find the second derivative of \( f(x) \):Start with the first derivative:\[ f'(x) = 12x - 3x^2 \]Differentiate again to find the second derivative:\[ f''(x) = \frac{d}{dx}(12x) - \frac{d}{dx}(3x^2) \]\[ f''(x) = 12 - 6x \].
04

Apply Second Derivative Test

For the critical points found, substitute into the second derivative:1. For \( x = 0 \):\[ f''(0) = 12 - 6(0) = 12 \]Since \( f''(0) > 0 \), \( x = 0 \) is a local minimum.2. For \( x = 4 \):\[ f''(4) = 12 - 6(4) = 12 - 24 = -12 \]Since \( f''(4) < 0 \), \( x = 4 \) is a local maximum.

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Key Concepts

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

First Derivative
The concept of the first derivative is crucial in finding critical points of a function. When we look for critical points, we start by taking the first derivative of the function, denoted as \( f'(x) \). This derivative helps us understand the rate of change or the slope of the function at various points. For the function \( f(x) = 9 + 6x^2 - x^3 \), the first derivative is calculated using the rules of differentiation.
  • The derivative of a constant, like \( 9 \), is \( 0 \).
  • The derivative of \( 6x^2 \) is \( 12x \).
  • The derivative of \( -x^3 \) is \( -3x^2 \).
Putting these together, we find \( f'(x) = 12x - 3x^2 \). The critical points are found where this derivative equals zero, leading to the equation \( 12x - 3x^2 = 0 \). Factoring out \( 3x \), the equation becomes \( 3x(4 - x) = 0 \), giving critical points at \( x = 0 \) and \( x = 4 \).
Second Derivative Test
After finding the critical points using the first derivative, the second derivative test helps to determine whether these points are local maxima, local minima, or points of inflection. This test involves taking the second derivative of the original function, denoted as \( f''(x) \). For \( f(x) = 9 + 6x^2 - x^3 \), we start from the first derivative \( f'(x) = 12x - 3x^2 \) and differentiate again:
  • The derivative of \( 12x \) is \( 12 \).
  • The derivative of \( -3x^2 \) is \( -6x \).
So, the second derivative is \( f''(x) = 12 - 6x \). By substituting the critical points into the second derivative, we can analyze the concavity of the function at these points. This is key in understanding whether a function is curving upwards or downwards at a critical point.
Local Maxima
Local maxima refer to points on the function where the function changes from increasing to decreasing, creating a peak. When using the second derivative test, if \( f''(x) < 0 \) at a critical point, this indicates a local maximum. This is because the negative sign shows the function is concave down at that point. In the exercise, the critical point \( x = 4 \) was tested with the second derivative:\[ f''(4) = 12 - 6(4) = -12 \]Since \( f''(4) < 0 \), this confirms that \( x = 4 \) is a local maximum. The function is curving downwards, indicating a peak at this point. Understanding local maxima is important for identifying the highest value in a specific interval of the domain.
Local Minima
Local minima are points where the function changes from decreasing to increasing, forming a valley. In the second derivative test, if \( f''(x) > 0 \) at a critical point, the function is concave up, indicating a local minimum. In the solution provided, the critical point \( x = 0 \) was analyzed using the second derivative:\[ f''(0) = 12 - 6(0) = 12 \]Since \( f''(0) > 0 \), this means \( x = 0 \) is a local minimum. The function is curving upwards at this point, forming a valley. Identifying local minima is key for determining the lowest value within a specified range of the function.

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