Choose the statement that is true for the graph f(x) =x^3-12x.
Answer: f(x) decreases for x >-2
2) Which type of critical point, if any, is present in the graph of f(x)=(x-4)^5+1
Answer: none
1) df/dx = 3x^2 -12 = 2(x+2)(x-2)
There are critical points (where the derivative is zero) at x = +2 and -2. x=2 is a relative minimum. f(x) increases (when x increases) for x > +2 and decreases for
-2 < x < 2
2) That equation has an absolute minimum at x = 4.
To determine if the statement "f(x) decreases for x > -2" is true for the graph of f(x) = x^3 - 12x, we need to understand the behavior of the function.
First, let's find the derivative of f(x) to identify the critical points. The derivative of f(x) can be found using the power rule:
f'(x) = 3x^2 - 12
To find the critical points, we set the derivative equal to zero and solve for x:
3x^2 - 12 = 0
Dividing both sides by 3:
x^2 - 4 = 0
Factoring the equation:
(x + 2)(x - 2) = 0
Setting each factor equal to zero:
x + 2 = 0 and x - 2 = 0
Solving for x, we get:
x = -2 and x = 2
Now, we can analyze the behavior of the function around these critical points. By evaluating the behavior of f(x) to the left and right of each critical point, we can determine whether the function increases or decreases in those regions.
For x < -2, we can pick a test point, such as x = -3, and substitute it into the function:
f(-3) = (-3)^3 - 12(-3) = -27 + 36 = 9
Considering x = -3 is to the left of the critical point x = -2, we can conclude that for x < -2, the function f(x) is increasing.
For values between -2 and 2, we can pick a test point, such as x = 0, and substitute it into the function:
f(0) = (0)^3 - 12(0) = 0
Considering x = 0 is within the interval -2 < x < 2, we can conclude that for -2 < x < 2, the function f(x) is constant.
For x > 2, we can pick a test point, such as x = 3, and substitute it into the function:
f(3) = (3)^3 - 12(3) = 27 - 36 = -9
Considering x = 3 is to the right of the critical point x = 2, we can conclude that for x > 2, the function f(x) is decreasing.
Therefore, the statement "f(x) decreases for x > -2" is true for the graph of f(x) = x^3 - 12x.
Now, let's move on to the second question.
For the function f(x) = (x-4)^5 + 1, we want to determine the type of critical point, if any, is present in the graph.
To find the critical points of f(x), we need to take the derivative of the function:
f'(x) = 5(x-4)^4
To find the critical points, we set the derivative equal to zero and solve for x:
5(x-4)^4 = 0
Since any number raised to the power of 4 is always positive, there are no values of x that make the derivative equal to zero. Therefore, there are no critical points in the graph of f(x) = (x-4)^5 + 1.
Hence, the answer to the second question is that there are no critical points present in the graph of f(x) = (x-4)^5 + 1.
To determine whether the statement "f(x) decreases for x > -2" is true for the graph of f(x) = x^3 - 12x, we can analyze the slope of the function.
1) Find the derivative of f(x) with respect to x:
f'(x) = 3x^2 - 12
2) Set the derivative equal to zero and solve for x to find critical points:
3x^2 - 12 = 0
3x^2 = 12
x^2 = 4
x = ±2
3) Determine the behavior of f(x) on the intervals (-∞, -2), (-2, 2), and (2, ∞):
- For x < -2, plug in a value less than -2 (e.g., -3) into f'(x): f'(-3) = 3(-3)^2 - 12 = 9 - 12 = -3
Since the derivative is negative, f(x) is decreasing for x < -2.
- For -2 < x < 2, plug in a value between -2 and 2 (e.g., 0) into f'(x): f'(0) = 3(0)^2 - 12 = -12
Since the derivative is negative, f(x) is also decreasing for -2 < x < 2.
- For x > 2, plug in a value greater than 2 (e.g., 3) into f'(x): f'(3) = 3(3)^2 - 12 = 27 - 12 = 15
Since the derivative is positive, f(x) is increasing for x > 2.
Therefore, the statement "f(x) decreases for x > -2" is false. The correct statement would be "f(x) decreases for x < -2 and -2 < x < 2".
Regarding the graph of f(x) = (x - 4)^5 + 1, we need to determine the type of critical points present, if any.
1) Find the derivative of f(x) with respect to x:
f'(x) = 5(x - 4)^4
2) Set the derivative equal to zero and solve for x to find critical points:
5(x - 4)^4 = 0
(x - 4)^4 = 0
(x - 4) = 0
x = 4
3) Determine the behavior of f(x) around x = 4:
Since the derivative is always positive or negative, the graph of f(x) = (x - 4)^5 + 1 does not have any critical points (neither local maxima nor minima). The graph is constantly increasing or decreasing without any turning points.
Therefore, the correct answer is "none" in terms of the critical points.