MTH1010_10_01_2

Question

MTH1010_10_01_2

BlackTom AI · Accepted Answer

We start by identifying the critical points of f by taking its derivative and setting it to zero. 
f'(x) = d/dx [2x^3 - 3x^2 - 12x + 11] = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1).
Critical points occur where f'(x) = 0, so x = 2 and x = -1.

Next, determine which of these is a local minimum or maximum using the second derivative test. The second derivative is f''(x) = d/dx [6x^2 - 6x - 12] = 12x - 6.
- At x = -1: f''(-1) = 12(-1) - 6 = -18 < 0, so this is a local maximum.
- At x = 2: f''(2) = 12(2) - 6 = 24 - 6 = 18 > 0, so this is a local minimum.

Now compute the y-value at the local minimum by evaluating f at x = 2:
f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 11 = 2(8) - 3(4) - 24 + 11 = 16 - 12 - 24 + 11 = -9.

With these results, we can assess each option:
Option a) asks for a particular x or y value that does not align with the computed critical points or their nature. The values here would not correspond to the local minimum position or its function value when checked against the derivative test and the second-derivative test.
Option b) similarly leads to inconsistencies with the derivative roots or the sign of the second derivative at those roots, so the classification of min vs max would be incorrect.
Option c) aligns with the findings: the local minimum occurs at x = 2 and the corresponding y-value is f(2) = -9. The calculation steps show x = 2 from the derivative roots and y = -9 from evaluating the polynomial at that x.
Option d) contains values that do not match either the critical points or the evaluation of f at those points, leading to an incorrect pair for the local minimum.

In summary, the local minimum is associated with the x-value from the positive second-derivative point and the y-value obtained by substituting that x into f, with the specific numeric result computed above.

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

MTH1010_10_01_2

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