Find all relative extrema of the function. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE:) f(x) = x3 6×2 + 6 relative maximum (x, Y) = ( relative minimum (x,Y) =
The Correct Answer and Explanation is:
To find the relative extrema of the function
f(x) = x³ − 6x² + 6,
we follow these steps:
Step 1: Find the critical points
Take the first derivative of the function:
f'(x) = d/dx [x³ − 6x² + 6] = 3x² − 12x
Set the first derivative equal to zero to find critical points:
3x² − 12x = 0
Factor:
3x(x − 4) = 0
So the critical points are:
x = 0 and x = 4
Step 2: Apply the Second Derivative Test
Find the second derivative:
f”(x) = d/dx [3x² − 12x] = 6x − 12
Evaluate the second derivative at each critical point.
- For x = 0:
f”(0) = 6(0) − 12 = −12
Since this is less than zero, the function is concave down at x = 0.
Therefore, there is a relative maximum at x = 0. - For x = 4:
f”(4) = 6(4) − 12 = 24 − 12 = 12
Since this is greater than zero, the function is concave up at x = 4.
Therefore, there is a relative minimum at x = 4.
Step 3: Find the function values at those points
- f(0) = 0³ − 6(0)² + 6 = 6
- f(4) = 4³ − 6(4)² + 6 = 64 − 96 + 6 = −26
Final Answer:
- Relative maximum: (x, y) = (0, 6)
- Relative minimum: (x, y) = (4, −26)
Explanation
To identify the relative extrema of a function, we look for critical points where the first derivative is zero or undefined. For the function f(x) = x³ − 6x² + 6, the first derivative is f'(x) = 3x² − 12x. Setting this equal to zero gives us the critical points: x = 0 and x = 4.
Next, we use the second derivative test to classify these critical points. The second derivative, f”(x) = 6x − 12, helps us determine concavity. If the second derivative is positive at a critical point, the graph is concave up, and the point is a relative minimum. If the second derivative is negative, the graph is concave down, and the point is a relative maximum.
Evaluating the second derivative at x = 0, we find f”(0) = −12, which is negative. This indicates that the function is concave down at x = 0, so this point is a relative maximum. At x = 4, f”(4) = 12, which is positive, showing that the function is concave up, and the point is a relative minimum.
Finally, we substitute these x-values into the original function to find their corresponding y-values. At x = 0, f(0) = 6, and at x = 4, f(4) = −26. So the relative maximum is at (0, 6), and the relative minimum is at (4, −26).
This analysis shows how calculus tools like the first and second derivatives help identify and classify relative extrema in polynomial functions, giving valuable information about the graph’s shape and turning points.
