f(x)=x^3-6x+1
find the critical points, where the function is increasing or decreasing and explain the shape of the graph using the derivative and algebra.
i found the derivative to be 3x^2-6. So the critical points are -1.4 and 1.4. The derivative is decreasing until (0,6) and then it begins increasing. i don't know how to go about explaining the shape of the graph with algebra though. is it related to quadratic equations???
To explain the shape of the graph using the derivative and algebra, we can analyze the sign of the derivative and the concavity of the function.
1. Finding the critical points:
To find the critical points, we set the derivative equal to zero and solve for x:
3x^2 - 6 = 0
Factoring, we get:
3(x^2 - 2) = 0
Setting each factor equal to zero: x^2 - 2 = 0
Solving for x, we find two critical points: x = √2 and x = -√2
2. Analyzing the sign of the derivative:
Using the derivative, 3x^2 - 6, we can determine where the function is increasing or decreasing.
To analyze the sign of the derivative, we pick test points in the intervals (-∞, -√2), (-√2, √2), and (√2, ∞) and evaluate the derivative at these points.
For x = -2:
3(-2)^2 - 6 = 12 > 0, meaning the derivative is positive in this interval.
For x = 0:
3(0)^2 - 6 = -6 < 0, meaning the derivative is negative in this interval.
For x = 2:
3(2)^2 - 6 = 6 > 0, meaning the derivative is positive in this interval.
3. Explaining the shape of the graph:
Based on the sign of the derivative, we can determine if the function is increasing or decreasing as well as identifying any local extremums (maximum or minimum) or points of inflection.
- In the interval (-∞, -√2), the derivative is positive, indicating that the function is increasing.
- At x = -√2, we have a minimum or local extremum point.
- In the interval (-√2, √2), the derivative is negative, indicating that the function is decreasing.
- At x = √2, we have a maximum or local extremum point.
- In the interval (√2, ∞), the derivative is positive, indicating that the function is increasing.
Regarding the shape of the graph, the function f(x) = x^3 - 6x + 1 is a cubic function. It could have a variety of shapes, but since we found points of local minimum and maximum, we know that the function has a point of inflection. The function starts decreasing, reaches a minimum at x = -√2, increases until it reaches a maximum at x = √2, and then continues to increase.
Note: When talking about the shape of the graph using algebra, we consider the sign of the derivative to determine if the function is increasing or decreasing and identify the presence of local extremums or points of inflection. Quadratic equations are related to the shape of the parabola, which is a 2nd-degree polynomial. In this case, the function is a 3rd-degree polynomial.
To find the critical points, we need to set the derivative of the function equal to zero and solve for x. Let's start with the derivative you found:
f'(x) = 3x^2 - 6
Setting f'(x) equal to zero, we have:
3x^2 - 6 = 0
To solve this quadratic equation, we can factor it:
3(x^2 - 2) = 0
Now, set each factor equal to zero and solve for x:
x^2 - 2 = 0
x^2 = 2
Taking the square root of both sides:
x = ±√2
So, the critical points are x = √2 and x = -√2 (approximately 1.41 and -1.41, respectively).
Once we have the critical points, we can determine whether the function is increasing or decreasing by analyzing the sign of the derivative in the intervals between and around the critical points.
To do this, we can choose test points in each interval and plug them into the derivative. Let's pick test points -2, 0, and 2:
For x < -√2 (e.g., x = -2):
f'(-2) = 3(-2)^2 - 6 = 12 > 0
For -√2 < x < √2 (e.g., x = 0):
f'(0) = 3(0)^2 - 6 = -6 < 0
For x > √2 (e.g., x = 2):
f'(2) = 3(2)^2 - 6 = 6 > 0
From the test points, we can conclude that the function is increasing before x = -√2, decreasing between x = -√2 and x = √2, and increasing again after x = √2.
To explain the shape of the graph using algebra, we can consider the behavior of the derivative. The derivative f'(x) = 3x^2 - 6 is a quadratic equation with a positive leading coefficient (3). This means that the graph of the derivative is an upward-opening parabola.
Since the derivative represents the rate of change of the function f(x), a positive value of the derivative indicates that the function is increasing, and a negative value indicates that the function is decreasing. Additionally, the points where the derivative is zero correspond to the critical points.
In our case, the derivative changes sign at x = -√2 and x = √2, which are the critical points. The graph of the function f(x) will have a local minimum at x = -√2 and a local maximum at x = √2. Since the function is increasing before -√2 and after √2, it must be decreasing between these points.
Therefore, the shape of the graph can be described as a curve that approaches a local minimum when x approaches -∞, then increases, reaches a local maximum at x = √2, decreases, reaches a local minimum at x = -√2, and then increases again as x approaches ∞.