Compute and present the first-order derivative f ′ (x). Discuss the interpretation of f ′ (x) and identify critical points.
11 answers
To compute the first-order derivative f'(x), we need the expression of the function f(x). Since it is not provided in the question, we cannot compute f'(x) or discuss its interpretation or identify critical points without that information. Please provide the function f(x) for further assistance.
f(x)=x^3−3x^2+2
To compute the first-order derivative f'(x) of f(x)=x^3−3x^2+2, we can apply the power rule of differentiation.
Using the power rule, the derivative of x^n, where n is a real number, is given by nx^(n-1).
So, applying the power rule to each term in the function, we obtain:
f'(x) = d/dx (x^3) - d/dx (3x^2) + d/dx(2)
f'(x) = 3x^2 - 2(3x) + 0
Simplifying the expression, we get:
f'(x) = 3x^2 - 6x
Now, let's discuss the interpretation of f'(x) and identify the critical points.
Interpretation of f'(x):
f'(x) represents the rate of change of the function f(x) with respect to x. In other words, it gives us the slope of the tangent line to the function at any given point.
Identifying critical points:
To identify the critical points, we need to find the x-values where the derivative f'(x) equals zero or is undefined.
Setting f'(x) = 0, we have:
3x^2 - 6x = 0
Factorizing the equation, we get:
3x(x - 2) = 0
This equation is satisfied when x = 0 or x = 2.
Therefore, the critical points of the function f(x) = x^3−3x^2+2 are x = 0 and x = 2.
Using the power rule, the derivative of x^n, where n is a real number, is given by nx^(n-1).
So, applying the power rule to each term in the function, we obtain:
f'(x) = d/dx (x^3) - d/dx (3x^2) + d/dx(2)
f'(x) = 3x^2 - 2(3x) + 0
Simplifying the expression, we get:
f'(x) = 3x^2 - 6x
Now, let's discuss the interpretation of f'(x) and identify the critical points.
Interpretation of f'(x):
f'(x) represents the rate of change of the function f(x) with respect to x. In other words, it gives us the slope of the tangent line to the function at any given point.
Identifying critical points:
To identify the critical points, we need to find the x-values where the derivative f'(x) equals zero or is undefined.
Setting f'(x) = 0, we have:
3x^2 - 6x = 0
Factorizing the equation, we get:
3x(x - 2) = 0
This equation is satisfied when x = 0 or x = 2.
Therefore, the critical points of the function f(x) = x^3−3x^2+2 are x = 0 and x = 2.
find second Derivative
To find the second derivative f''(x), we differentiate f'(x) = 3x^2 - 6x using the power rule once again:
f''(x) = d/dx (3x^2 - 6x)
Apply the power rule, we get:
f''(x) = 6x - 6
So, the second derivative of f(x) = x^3−3x^2+2 is f''(x) = 6x - 6.
f''(x) = d/dx (3x^2 - 6x)
Apply the power rule, we get:
f''(x) = 6x - 6
So, the second derivative of f(x) = x^3−3x^2+2 is f''(x) = 6x - 6.
Compute and present the second-order derivative � ′ ′ ( � ) f ′′ (x). Discuss the interpretation of � ′ ′ ( � ) f ′′ (x) and its relationship to concavity.
To compute the second-order derivative f''(x), we first need the first derivative f'(x), which we previously found to be f'(x) = 3x^2 - 6x.
Now, let's find the second-order derivative f''(x):
f''(x) = d/dx (3x^2 - 6x)
Using the power rule, we get:
f''(x) = 6x - 6
So, the second-order derivative of f(x) = x^3−3x^2+2 is f''(x) = 6x - 6.
Interpretation of f''(x) and its relationship to concavity:
The second-order derivative f''(x) provides information about the concavity of the function f(x).
If f''(x) > 0, the function is concave up, meaning the graph is shaped like a "U" or a smiling face. In this case, the function is increasing at an increasing rate.
If f''(x) < 0, the function is concave down, meaning the graph is shaped like an upside-down "U" or a frowning face. In this case, the function is increasing at a decreasing rate.
In the case of f''(x) = 6x - 6, since the coefficient of x (6) is positive, f''(x) is greater than zero, indicating that the function is concave up. This means that the graph of f(x) is shaped like a "U" and the function is increasing at an increasing rate.
Now, let's find the second-order derivative f''(x):
f''(x) = d/dx (3x^2 - 6x)
Using the power rule, we get:
f''(x) = 6x - 6
So, the second-order derivative of f(x) = x^3−3x^2+2 is f''(x) = 6x - 6.
Interpretation of f''(x) and its relationship to concavity:
The second-order derivative f''(x) provides information about the concavity of the function f(x).
If f''(x) > 0, the function is concave up, meaning the graph is shaped like a "U" or a smiling face. In this case, the function is increasing at an increasing rate.
If f''(x) < 0, the function is concave down, meaning the graph is shaped like an upside-down "U" or a frowning face. In this case, the function is increasing at a decreasing rate.
In the case of f''(x) = 6x - 6, since the coefficient of x (6) is positive, f''(x) is greater than zero, indicating that the function is concave up. This means that the graph of f(x) is shaped like a "U" and the function is increasing at an increasing rate.
Local and Global Extrema
To find the local and global extrema of a function, we need to find the critical points and analyze the behavior of the function at these points as well as at the endpoints of the domain.
For the function f(x) = x^3 - 3x^2 + 2:
1. Critical points:
To find the critical points, we set the first derivative f'(x) equal to zero and solve for x:
f'(x) = 3x^2 - 6x = 0
Factorizing the equation, we get:
3x(x - 2) = 0
This equation is satisfied when x = 0 or x = 2.
Therefore, the critical points are x = 0 and x = 2.
2. Endpoints:
Since the given function f(x) = x^3 - 3x^2 + 2 is an algebraic expression, it is defined for all real numbers. Thus, there are no endpoints to consider.
Now, we need to analyze the behavior of the function at the critical points and determine if they correspond to local or global extrema.
At x = 0:
To determine the nature of the critical point at x = 0, we analyze the sign changes of the first derivative f'(x) on either side of 0.
f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 (positive)
f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 (negative)
Since there is a sign change from positive to negative around x = 0, the function has a local maximum at x = 0.
At x = 2:
Similarly, analyzing the signs of f'(x) on either side of 2:
f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 (negative)
f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 (positive)
There is a sign change from negative to positive around x = 2, indicating a local minimum at x = 2.
As for the global extrema of the function f(x), since the function is a cubic polynomial and has no other critical points between the given critical points, it means that the local extrema at x = 0 and x = 2 are also the global extrema for the function.
Therefore, the function f(x) = x^3 - 3x^2 + 2 has a local maximum at x = 0 and a local minimum at x = 2, and both of these points are also the global extrema of the function.
For the function f(x) = x^3 - 3x^2 + 2:
1. Critical points:
To find the critical points, we set the first derivative f'(x) equal to zero and solve for x:
f'(x) = 3x^2 - 6x = 0
Factorizing the equation, we get:
3x(x - 2) = 0
This equation is satisfied when x = 0 or x = 2.
Therefore, the critical points are x = 0 and x = 2.
2. Endpoints:
Since the given function f(x) = x^3 - 3x^2 + 2 is an algebraic expression, it is defined for all real numbers. Thus, there are no endpoints to consider.
Now, we need to analyze the behavior of the function at the critical points and determine if they correspond to local or global extrema.
At x = 0:
To determine the nature of the critical point at x = 0, we analyze the sign changes of the first derivative f'(x) on either side of 0.
f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 (positive)
f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 (negative)
Since there is a sign change from positive to negative around x = 0, the function has a local maximum at x = 0.
At x = 2:
Similarly, analyzing the signs of f'(x) on either side of 2:
f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 (negative)
f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 (positive)
There is a sign change from negative to positive around x = 2, indicating a local minimum at x = 2.
As for the global extrema of the function f(x), since the function is a cubic polynomial and has no other critical points between the given critical points, it means that the local extrema at x = 0 and x = 2 are also the global extrema for the function.
Therefore, the function f(x) = x^3 - 3x^2 + 2 has a local maximum at x = 0 and a local minimum at x = 2, and both of these points are also the global extrema of the function.
dentify intervals where the function is concave up and concave down.
Explain the significance of these intervals in understanding the shape of the graph.
Explain the significance of these intervals in understanding the shape of the graph.
To identify the intervals in which the function f(x) = x^3 - 3x^2 + 2 is concave up or concave down, we need to analyze the second derivative f''(x) = 6x - 6.
Concave Up:
If f''(x) > 0, the function is concave up. In other words, the graph of the function is shaped like a "U".
Considering f''(x) = 6x - 6 > 0, we can solve for x:
6x - 6 > 0
6x > 6
x > 1
So, f(x) is concave up for x > 1.
Concave Down:
If f''(x) < 0, the function is concave down. In other words, the graph of the function is shaped like an upside-down "U".
Considering f''(x) = 6x - 6 < 0, we can solve for x:
6x - 6 < 0
6x < 6
x < 1
So, f(x) is concave down for x < 1.
Significance of Concavity in Understanding the Graph:
Understanding the concavity of a function is crucial for understanding the shape of its graph. It helps us identify the points of inflection, where the concavity changes.
In the case of f(x) = x^3 - 3x^2 + 2, we have found that the function is concave up for x > 1 and concave down for x < 1. This means that at x = 1, there is a point of inflection where the concavity changes from concave down to concave up.
Knowing where the function is concave up or concave down helps us visualize the shape of the graph more accurately. For example, if we only knew that the function is increasing or decreasing, we wouldn't have a clear understanding of the curvature of the graph. The concavity provides additional information about the rate at which the function is changing, allowing us to better interpret and analyze its behavior.
Concave Up:
If f''(x) > 0, the function is concave up. In other words, the graph of the function is shaped like a "U".
Considering f''(x) = 6x - 6 > 0, we can solve for x:
6x - 6 > 0
6x > 6
x > 1
So, f(x) is concave up for x > 1.
Concave Down:
If f''(x) < 0, the function is concave down. In other words, the graph of the function is shaped like an upside-down "U".
Considering f''(x) = 6x - 6 < 0, we can solve for x:
6x - 6 < 0
6x < 6
x < 1
So, f(x) is concave down for x < 1.
Significance of Concavity in Understanding the Graph:
Understanding the concavity of a function is crucial for understanding the shape of its graph. It helps us identify the points of inflection, where the concavity changes.
In the case of f(x) = x^3 - 3x^2 + 2, we have found that the function is concave up for x > 1 and concave down for x < 1. This means that at x = 1, there is a point of inflection where the concavity changes from concave down to concave up.
Knowing where the function is concave up or concave down helps us visualize the shape of the graph more accurately. For example, if we only knew that the function is increasing or decreasing, we wouldn't have a clear understanding of the curvature of the graph. The concavity provides additional information about the rate at which the function is changing, allowing us to better interpret and analyze its behavior.