Question What is the relative maximum and minimum of the function? = 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).

Question

Question What is the relative maximum and minimum of the function? = 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).

Answer 1

The relative maximum is at (–1.53, 12.01) and the relative minimum is at (1.2, –8.3).

Answer 2

The correct answer is:

The relative maximum is at (-1.2, 8.3) and the relative minimum is at (1.53, -12.01).

Answer 3

To find the relative maximum and minimum of a function, you need to find the critical points and perform the second derivative test.

First, let's find the critical points by taking the derivative of the function:

f(x) = 2x^3 + x^2 - 11x

f'(x) = 6x^2 + 2x - 11

To find the critical points, we set f'(x) equal to zero and solve for x:

6x^2 + 2x - 11 = 0

We can solve this quadratic equation using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, a = 6, b = 2, and c = -11. Plugging these values into the quadratic formula, we get:

x = (-2 ± sqrt(2^2 - 4(6)(-11))) / (2(6))
x = (-2 ± sqrt(4 + 264)) / 12
x = (-2 ± sqrt(268)) / 12
x = (-2 ± 16.37) / 12

This gives us two possible values for x: x = (-2 + 16.37) / 12 ≈ 1.20 and x = (-2 - 16.37) / 12 ≈ -1.53.

Now, to determine whether these critical points are relative maxima or minima, we need to perform the second derivative test. We take the second derivative of the function:

f''(x) = 12x + 2

We evaluate f''(x) at each of the critical points we found:

f''(1.20) ≈ 14.40
f''(-1.53) ≈ -16.36

If f''(x) is positive at a critical point, it indicates a relative minimum, and if f''(x) is negative, it indicates a relative maximum.

So, we can conclude that the relative maximum is at approximately (-1.53, 12.01) and the relative minimum is at approximately (1.20, -8.3).

Therefore, the correct response is:
"The relative maximum is at (–1.53, 12.01) and the relative minimum is at (1.20, –8.3)."