(A-Level) Find the coordinate of the stationary point of the curve y = 2x + 27/x^2

Since we are looking for a stationary this means the derivative will be equal to 0, so we will have to differentiate the equation. When we differentiate ( y = 2x + 27/x^2 ) we get ( dy/dx = 2 - 54/x^3 ). Because we are looking for a stationary point we set this equal to 0 and solve ( 2 - 54/x^3 = 0), rearranging this we get ( 2 = 54/x^3 ), multipying both sides by x^3 we get ( 2x^3 = 54 ), dividing this by 2, ( x^3 = 27 ), and now taking the cube root of both sides we get ( x = 3 ). So we know that the x-coordinate of the stationary point is 3, we can plug this back into the original equation to find the y-coordinate. Doing this we get ( y = 2*3 + 27/9 = 6 + 3 = 9 ), so the y-coordinate is 9. Thus the coordinate of the stationary point is (3,9)