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