What is the relative maximum and minimum of the function?
function symbol= 2x3 + x2 – 11x
(1 point)
Responses
The relative maximum is at (–1.53, 8.3) and the relative minimum is at (1.2, –12.01).
The relative maximum is at (–1.53, 8.3) and the relative minimum is at (1.2, –12.01).
The relative maximum is at (–1.53, 12.01) and the relative minimum is at (1.2, –8.3).
The relative maximum is at (–1.53, 12.01) and the relative minimum is at (1.2, –8.3).
The relative maximum is at (–1.2, 8.3) and the relative minimum is at (1.53, –12.01).
The relative maximum is at (–1.2, 8.3) and the relative minimum is at (1.53, –12.01).
The relative maximum is at (–1.2, 12.01) and the relative minimum is at (1.53, –8.3).
The relative maximum is at (–1.2, 12.01) and the relative minimum is at (1.53, –8.3).
To determine the relative maximum and minimum of a function, follow these steps:
1. Take the derivative of the function to find critical points. In this case, the derivative of the function will be:
f'(x) = 6x^2 + 2x - 11
2. Set the derivative equal to zero and solve for x to find the critical points:
6x^2 + 2x - 11 = 0
3. Solve the quadratic equation using factoring, the quadratic formula, or any other method and find the values of x.
4. Once you have the values of x, substitute them back into the original function to find the corresponding y-values.
5. Compare the y-values to determine the relative maximum and minimum. The highest y-value will be the relative maximum, and the lowest y-value will be the relative minimum.
Using these steps, you can find the values of x and y to determine the correct answer from the options provided.
To find the relative maximum and minimum of the function f(x) = 2x^3 + x^2 - 11x, we need to take the derivative of the function and set it equal to zero to find the critical points. Then we can determine whether each critical point is a relative maximum or minimum by examining the sign of the second derivative.
1. Find the derivative of f(x):
f'(x) = 6x^2 + 2x - 11
2. Set f'(x) = 0 and solve for x to find the critical points:
6x^2 + 2x - 11 = 0
You can solve this quadratic equation using the quadratic formula or factoring. Either way, you will find two solutions for x: approximately -1.53 and 1.2.
3. Find the second derivative of f(x):
f''(x) = 12x + 2
4. Evaluate the second derivative at each critical point:
f''(-1.53) = -17.36
f''(1.2) = 14.4
5. Determine the nature of each critical point based on the sign of the second derivative:
Since f''(-1.53) is negative, the function has a relative maximum at x = -1.53.
Since f''(1.2) is positive, the function has a relative minimum at x = 1.2.
Therefore, the relative maximum is at (-1.53, 8.3) and the relative minimum is at (1.2, -12.01). The correct response is:
The relative maximum is at (-1.53, 8.3) and the relative minimum is at (1.2, -12.01).
Miguel is designing shipping boxes that are rectangular prisms. The shape of one box, with height h in feet, has a volume defined by the function V(h) = h(–h + 10)(–h + 8). What is the maximum volume for the domain
0 < h < 10? Round to the nearest cubic foot.
(1 point)
Responses
10 ft3
10 ft 3
107 ft3
107 ft 3
105 ft3
105 ft 3
110 ft3
To find the maximum volume for the domain 0 < h < 10, we need to find the critical points of the volume function V(h) = h(-h + 10)(-h + 8) and determine which one corresponds to the maximum value.
1. Take the derivative of the volume function:
V'(h) = -h^2 + 18h - 80
2. Set the derivative equal to zero and solve for h to find the critical points:
-h^2 + 18h - 80 = 0
3. Solve the quadratic equation using factoring, the quadratic formula, or any other method and find the values of h. In this case, the quadratic equation doesn't factor nicely, so we can use the quadratic formula:
h = (-b ± √(b^2 - 4ac)) / (2a)
Using a = -1, b = 18, and c = -80, we get:
h = (-18 ± √(18^2 - 4(-1)(-80))) / (2(-1))
h = (-18 ± √(324 - 320)) / (-2)
h = (-18 ± √4) / (-2)
h = (-18 ± 2) / (-2)
Simplifying further:
h = (-18 + 2) / (-2) or h = (-18 - 2) / (-2)
h = -16 / -2 or h = -20 / -2
h = 8 or h = 10
4. Check if the critical points are within the given domain of 0 < h < 10.
Since h = 8 is within the domain, we can continue with this critical point.
5. Evaluate the volume function at the critical point and the endpoints of the given domain:
V(0) = 0(-0 + 10)(-0 + 8) = 0
V(8) = 8(-8 + 10)(-8 + 8) = 0
V(10) = 10(-10 + 10)(-10 + 8) = 0
Since the volume at h = 0, h = 8, and h = 10 are all 0, the maximum volume must occur at the critical point within the domain.
Therefore, the maximum volume for the given domain 0 < h < 10 is 0 ft^3. The correct response is:
10 ft3 (incorrect)
107 ft3 (incorrect)
105 ft3 (incorrect)
110 ft3 (incorrect)
None of the options provided, including the given domain, satisfy the conditions of the problem.