Given that y = 16x + x^-1, find the two values of x for which dy/dx = 0
Expected layout of answer in an exam:y = 16x + x-1dy/dx = 16 - x-2dy/dx = 0 implies that 16 - x-2 = 0this implies that 16 = x-216 = 1/x216x2 = 1x2 = 1/16therefore x = +/- 1/4x = 1/4, x = -1/4 when dy/dx = 0
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