find the coordinates of the single stationary point of the curve with equation y=8x^2 + 2/x
rewrite the equation in a more helpful way:y=8x2 + 2x-1 differentiate:dy/dx = 16x -2x-2set equal to 0 and solve for x16x -2x-2 = 016x -2/ x2=016x= 2/ x216x3=28x3=1x3=1/8x=1/2note: x cannot be -1/2sub x into original curve equation:y= 8(0.5)2 + 2/0.5y= 2+ 4y=6
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