Maths differentiation question

Given that y=16x+x^-1, find the two values of x for which dy/dx =0.
Thanks :)

y=16x+x^-1
dy/dx = 16 - x^(-2)

when dy/dx = 0

[0 = 16 - 1/x^2] x^2
0 = 16x^2 - 1
1 = 16x^2
x^2 = 1/16
x = ± √(1/16)
x = ± 1/4

x = 1/4, - 1/4 answer//

y=16x+x^-1

y’ = 16 – 1/x² = 0

So 1/x² = 16 or x² = 1/16

so x = ¼ or –¼

Yeah, Algebra is everywhere!

y=16x+x^-1
dy/dx = 16 +(-1)x^(-2)
dy/dx = 16-1/x^2 = 0
16=1/x^2
x^2 = 1/16
x = ± 1/4

dy/dx=16-1/x^2=0
16x^2=1

16x^2-1=0
4x+1(4x-1)=0
x=-1/4
x=1/4
Is that okaY?