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Consider f(x) = x⁵ - 20*x³ + 50*x + 2. Identify the correct set of intervals where f(x) is strictly decreasing.

1. (-3.32, -0.95) and (0.95, 3.32)
2. (-infinity, -3.32) and (-0.95, 0.95)
3. (-infinity, -3.32) and (0.95, 3.32)
4. (-3.32, -0.95) and (0.95, infinity)

Correct answer: (-3.32, -0.95) and (0.95, 3.32)

Solution

f'(x) = 5*x⁴ - 60*x² + 50 = 5*(x⁴ - 12*x² + 10). Let t = x²: t² - 12t + 10 = 0, giving t = (12 +/- sqrt(144-40))/2 = (12 +/- sqrt(104))/2 = 6 +/- sqrt(26). sqrt(26) approx 5.099. So t1 = 6 - 5.099 = 0.901 (x = +/-0.949 approx +/-0.95) and t2 = 6 + 5.099 = 11.099 (x = +/-3.332 approx +/-3.33). f'(x) is a degree-4 polynomial with positive leading coefficient, so it is positive for |x| large and negative between consecutive roots. The sign pattern: positive on (-inf,-3.33), negative on (-3.33,-0.95), positive on (-0.95,0.95), negative on (0.95,3.33), positive on (3.33,inf). So f is decreasing on (-3.33,-0.95) and (0.95,3.33).

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