To find the relative maximum and minimum of the function f(x) = 2x³ + x² – 11x, we first need to find its critical points. These are the points where the derivative of the function is either zero or undefined.
First, we find the derivative of the function:
f'(x) = 6x² + 2x – 11
Next, we set the derivative equal to zero to find the critical points:
6x² + 2x – 11 = 0
We can solve this quadratic equation using the quadratic formula: x = (-b ± √(b² – 4ac)) / 2a, where a = 6, b = 2, and c = -11.
Plugging in these values:
x = ( -2 ± √(2² – 4 * 6 * -11) ) / (2 * 6)
x = ( -2 ± √(4 + 264) ) / 12
x = ( -2 ± √268 ) / 12
x = ( -2 ± 2√67 ) / 12
x = (-1 ± √67) / 6
This gives us two critical points:
x1 = (-1 + √67) / 6 and x2 = (-1 – √67) / 6
Now, to determine whether these critical points are relative maxima or minima, we will use the second derivative test. We first find the second derivative:
f”(x) = 12x + 2
We then evaluate the second derivative at our critical points.
For x1 = (-1 + √67) / 6:
f”(x1) = 12((-1 + √67) / 6) + 2
If this value is positive, it indicates a relative minimum, and if it’s negative, it indicates a relative maximum.
For x2 = (-1 – √67) / 6:
f”(x2) = 12((-1 – √67) / 6) + 2
By evaluating f” at both critical points, we can classify each point accordingly:
- If f”(x1) > 0, then there is a relative minimum at x1.
- If f”(x2) < 0, then there is a relative maximum at x2.
In summary, the relative maximum and minimum can be determined by following these steps: finding the derivative, solving for critical points, and using the second derivative for classification. This process gives us insights into the behavior of the function and where it peaks or dips.