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If x = 1 is a critical point of the function f(x) = (3x² + ax - 2 - a)e^x, then:
- x = 1 is a local minima and x = -2/3 is a local maxima of f
- x = 1 is a local maxima and x = -2/3 is a local minima of f
- x = 1 and x = -2/3 are local minima of f
- x = 1 and x = -2/3 are local maxima of f
Correct answer: x = 1 is a local minima and x = -2/3 is a local maxima of f
Solution
f'(x) = (6x + a + 3x^2 + ax - 2 - a)e^x. Setting f'(1) = 0 gives a = -7. Then the critical points are x = 1 and x = -2/3. f''(1) = 5e > 0 (local minimum) and f''(-2/3) = -5e^(-2/3) < 0 (local maximum). So x = 1 is a local minimum and x = -2/3 is a local maximum.
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