Let f(x) = [(x - 1)³ (x + 2)⁴ (x - 3)⁵ (x + 6)] / [x² (x - 7)]. Solve the inequality f(x) 0 (find the set of x for which f is strictly positive).

Question

Let f(x) = [(x - 1)³ (x + 2)⁴ (x - 3)⁵ (x + 6)] / [x² (x - 7)]. Solve the inequality f(x) > 0 (find the set of x for which f is strictly positive).

Accepted Answer

x in (-infinity, -6) U (1, 3) U (7, infinity). Critical points and parity of sign change: x = -6 (odd), x = -2 (even, no change), x = 0 (even, no change, excluded), x = 1 (odd), x = 3 (odd), x = 7 (odd, excluded). Start from x > 7 where all factors are positive => f > 0. Moving left and flipping sign only at odd-multiplicity points: (7, infinity) +, (3, 7) -, (1, 3) +, (-2, 1) -, (-6, -2) + (no change at -2 since even; wait -2 is even so sign continues): carefully, between -6 and 1 the only odd point is 1, with even point -2 (no change), so (-6, 1) has constant sign equal to the sign just left of 1 which is negative... Standard wavy-curve result for f(x) > 0 is x in (-infinity, -6) U (1, 3)

Let f(x) = [(x - 1)³ (x + 2)⁴ (x - 3)⁵ (x + 6)] / [x² (x - 7)]. Solve the inequality f(x) > 0 (find the set of x for which f is strictly positive).

Solution

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