Answer:
To find the minimum and maximum of the function -4x^3 - 5x^2 + 10x - 2, we can follow these steps:
1. Take the derivative of the function to find critical points:
f(x) = -4x^3 - 5x^2 + 10x - 2
f'(x) = -12x^2 - 10x + 10
2. Set the derivative equal to zero and solve for x to find the critical points:
-12x^2 - 10x + 10 = 0
3. Solve the quadratic equation by factoring or using the quadratic formula. In this case, factoring is not possible, so let's use the quadratic formula:
x = (-(-10) ± √((-10)^2 - 4(-12)(10))) / (2(-12))
Simplifying the expression inside the square root:
x = (-(-10) ± √(100 + 480)) / (-24)
x = (10 ± √580) / (-24)
4. Calculate the values of x by using the positive and negative solutions from the quadratic formula:
x ≈ 0.5486 or x ≈ -1.9653
5. Now, we need to determine whether these critical points correspond to a minimum or maximum. To do this, we can use the second derivative test. Take the second derivative of the function:
f''(x) = -24x - 10
6. Substitute the critical points into the second derivative:
For x ≈ 0.5486:
f''(0.5486) = -24(0.5486) - 10 ≈ -22.367
Since the second derivative is negative (-22.367 < 0), this indicates a maximum at x ≈ 0.5486.
For x ≈ -1.9653:
f''(-1.9653) = -24(-1.9653) - 10 ≈ 38.767
Since the second derivative is positive (38.767 > 0), this indicates a minimum at x ≈ -1.9653.
7. So, the minimum occurs at x ≈ -1.9653, and the maximum occurs at x ≈ 0.5486.
The minimum value of the function is obtained by substituting x ≈ -1.9653 into the original function:
f(-1.9653) = -4(-1.9653)^3 - 5(-1.9653)^2 + 10(-1.9653) - 2 ≈ -22.318
The maximum value of the function is obtained by substituting x ≈ 0.5486 into the original function:
f(0.5486) = -4(0.5486)^3 - 5(0.5486)^2 + 10(0.5486) - 2 ≈ 1.977
Therefore, the minimum value is approximately -22.318, and the maximum value is approximately 1.977.