What is the relative maximum and minimum of the function?
function symbol= 2x^3 + x^2 – 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, 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, 12.01) and the relative minimum is at (1.53, –8.3).
The relative maximum is at (–1.2, 8.3) and the relative minimum is at (1.53, –12.01).
To find the relative maximum and minimum of the function, we need to find the critical points.
First, we find the derivative of the function:
f'(x) = 6x^2 + 2x - 11
Next, we set the derivative equal to zero to find the critical points:
6x^2 + 2x - 11 = 0
We can use the quadratic formula to solve for x:
x = (-b ± √(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 ± √(2^2 - 4(6)(-11))) / (2(6))
x = (-2 ± √(4 + 264)) / 12
x = (-2 ± √268) / 12
x ≈ -1.53 or x ≈ 1.2
Now, we need to find the corresponding y-values for these x-values:
f(-1.53) ≈ 8.3
f(1.2) ≈ -12.01
Therefore, the relative maximum is at approximately (-1.53, 8.3) and the relative minimum is at approximately (1.2, -12.01).
Therefore, the correct answer is:
The relative maximum is at (–1.53, 8.3) and the relative minimum is at (1.2, –12.01).
To find the relative maximum and minimum of the function, you need to find the critical points of the function by taking its derivative and setting it equal to zero.
First, find the derivative of the function:
f'(x) = 6x^2 + 2x - 11
Now, set the derivative equal to zero and solve for x:
6x^2 + 2x - 11 = 0
Using the quadratic formula, you can find the roots of the equation to locate the critical points:
x = (-2 ± √(2^2 - 4(6)(-11))) / (2(6))
After solving this equation, you will find two critical points: x = -1.53 and x = 1.2.
To determine whether these points correspond to a relative maximum or minimum, you can analyze the concavity of the function.
Find the second derivative of the function:
f''(x) = 12x + 2
Now, substitute the critical points into the second derivative and evaluate the sign:
For x = -1.53:
f''(-1.53) = 12(-1.53) + 2 = -16.36
Since the second derivative is negative, this corresponds to a relative maximum.
For x = 1.2:
f''(1.2) = 12(1.2) + 2 = 16.4
Since the second derivative is positive, this corresponds to a relative minimum.
Therefore, the correct answer is:
The relative maximum is at (–1.53, 8.3) and the relative minimum is at (1.2, –12.01).