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Magazine Chapter 4: Problem 12
Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. \(f(x)=x^{3}-3 x+10\)
Short Answer
Expert verified
Critical points are at \(x = -1\) (max) and \(x = 1\) (min); inflection at \(x = 0\).
Step by step solution
01
Understanding Critical Points
Critical points occur where the first derivative of a function is zero or undefined. For the function \( f(x) = x^3 - 3x + 10 \), we first need to take its derivative to find the critical points.
02
Calculate the First Derivative
Find the first derivative of the function \( f(x) = x^3 - 3x + 10 \). The derivative is given by \( f'(x) = 3x^2 - 3 \).
03
Find Critical Points
Set the first derivative equal to zero to find critical points: \( 0 = 3x^2 - 3 \). Solving this equation, we have \( x^2 = 1 \), so \( x = \pm 1 \). Thus, the critical points are at \( x = 1 \) and \( x = -1 \).
04
Calculate the Second Derivative
To analyze the nature of these critical points, we calculate the second derivative: \( f''(x) = 6x \).
05
Use the Second Derivative Test
Evaluate the second derivative at each critical point:- At \( x = 1 \): \( f''(1) = 6 \times 1 = 6 > 0 \), so \( x = 1 \) is a local minimum.- At \( x = -1 \): \( f''(-1) = 6 \times (-1) = -6 < 0 \), so \( x = -1 \) is a local maximum.
06
Determine Inflection Points
Inflection points occur where the second derivative changes sign. Set \( f''(x) = 6x = 0 \) to find potential inflection points, which gives \( x = 0 \). This is an inflection point since the sign of \( f''(x) \) changes around \( x = 0 \).
07
Graph and Validate Results
Plot the function \( f(x) = x^3 - 3x + 10 \) to check the positions of the local maximum at \( x = -1 \) and the local minimum at \( x = 1 \), and to observe the inflection point at \( x = 0 \). The graph will show a peak at \( x = -1 \) and a trough at \( x = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
The first derivative of a function is a powerful tool used to find critical points. A critical point is where the slope of the function is zero or undefined, which often indicates a point of interest like a peak or valley on the graph.
To find the first derivative of the function \( f(x) = x^3 - 3x + 10 \), we use differentiation. The derivative \( f'(x) \) is \( 3x^2 - 3 \).
Setting this equal to zero helps us find critical points:
To find the first derivative of the function \( f(x) = x^3 - 3x + 10 \), we use differentiation. The derivative \( f'(x) \) is \( 3x^2 - 3 \).
Setting this equal to zero helps us find critical points:
- \( 3x^2 - 3 = 0 \)
- \( x^2 = 1 \)
- This gives the solutions \( x = \pm 1 \).
Second Derivative
The second derivative gives us insight into the concavity of a function and is instrumental in determining the nature of critical points. When calculating it, we simply differentiate the first derivative. For our function, the first derivative \( f'(x) = 3x^2 - 3 \) gives us a second derivative \( f''(x) = 6x \).
This second derivative helps us understand how the slope of the function changes:
This second derivative helps us understand how the slope of the function changes:
- If \( f''(x) > 0 \), the function is concave up at that point.
- If \( f''(x) < 0 \), the function is concave down at that point.
Inflection Points
Inflection points occur where the second derivative changes sign. These are crucial because they indicate a change in the concavity of the function. The process involves setting the second derivative equal to zero and examining the sign changes around these points.
For our example, \( f''(x) = 6x \), which we set to zero:
For our example, \( f''(x) = 6x \), which we set to zero:
- \( 6x = 0 \)
- \( x = 0 \)
Local Maximum
A local maximum occurs at a critical point where the function reaches a peak, higher than any nearby points. To determine if we have a local maximum, we use the second derivative test.
According to our calculations, at the critical point \( x = -1 \), the second derivative \( f''(-1) = -6 \). Since this is less than zero, the function is concave down at this point, confirming that \( x = -1 \) is a local maximum.
Here, the graph will display a peak, which indicates the highest point in this neighborhood.
According to our calculations, at the critical point \( x = -1 \), the second derivative \( f''(-1) = -6 \). Since this is less than zero, the function is concave down at this point, confirming that \( x = -1 \) is a local maximum.
Here, the graph will display a peak, which indicates the highest point in this neighborhood.
Local Minimum
Conversely, a local minimum is present at a critical point if it is lower than all nearby points. The second derivative test helps us identify this situation.
For \( x = 1 \), we calculate the second derivative \( f''(1) = 6 \), which is greater than zero. This suggests the function is concave up at this point, confirming a local minimum.
At \( x = 1 \), the graph will illustrate a valley, marking the lowest point in this vicinity. Thus, understanding local minima and maxima through this method is crucial for analyzing functions.
For \( x = 1 \), we calculate the second derivative \( f''(1) = 6 \), which is greater than zero. This suggests the function is concave up at this point, confirming a local minimum.
At \( x = 1 \), the graph will illustrate a valley, marking the lowest point in this vicinity. Thus, understanding local minima and maxima through this method is crucial for analyzing functions.
