Question text 1. Answer 1 Question 10[input] 2. Answer 2 Question 10[input]

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The problem presents a function f(x) = 2x^3 − 3x^2 − 12x + 11 and asks for the x-value and y-value of its local minimum, with the x-value to be written as an integer in the first box and the y-value as an integer in the second box.

First, find the critical points by differentiating: f'(x) = 6x^2 − 6x − 12 = 6(x^2 − x − 2) = 6(x − 2)(x + 1). So the critical points occur at x = −1 and x = 2.

Next, determine which is a local maximum and which is a local minimum using the second derivative or first-derivative test. The second derivative is f''(x) = 12x − 6.
- At x = −1: f''(−1) = −18 < 0, so this is a local maximum (not the local minimum we’re asked for).
- At x = 2: f''(2) = 18 > 0, so this is a local minimum.

Now evaluate the function at the local minimum to obtain the corresponding y-value:
f(2) = 2(2)^3 − 3(2)^2 − 12(2) + 11 = 16 − 12 − 24 + 11 = −9.

Therefore, the x-value of the local minimum is 2, and the corresponding y-value is −9.

Why the provided second value (−13) is not correct: plugging x = 2 into f(x) yields −9, not −13. The y-value must match the actual function value at the x-coordinate of the local minimum.

Summary:
- x-value (local minimum): 2
- y-value (local minimum): −9

If you intended to input different numbers, re-check the substitution into f(x) at x = 2 to confirm the resulting y-value.

MCD2130 - T2 - 2025 11.7 Online Quiz - MCD2130_QUIZ_TEN Differentiation and Integration

Question text 1. Answer 1 Question 10[input] 2. Answer 2 Question 10[input]

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