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c2 differentiation

Find the coordinates of the stationary point on the graph of

y=x2+16x y= x^{2}+\frac{16}{x}

My answer is (2,12)

but the book says (2,8)

my working

dydx=2x−16x−2 \frac{\mathrm{d} y}{\mathrm{d} x}= 2x -16x^{-2}
let dy/dx=0
dydx=2x−16x−2=0 \frac{\mathrm{d} y}{\mathrm{d} x}= 2x -16x^{-2} =0

2x=16x22x=\frac{16}{x^{2}}

x3=8 x^{3}=8
x=2 x=2

x-coordinate of stationary point = 2
sub x=2 into y to find the y-coordinate of stationary point

x=2 x=2

y=(2)2+162=12 y= (2)^{2} +\frac{16}{2} = 12


is there any mistake?
Original post by bigmansouf
Find the coordinates of the stationary point on the graph of

y=x2+16x y= x^{2}+\frac{16}{x}

My answer is (2,12)

but the book says (2,8)

is there any mistake?


Agree with (2,12). Book's wrong.

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