Question

find the points where the tangent to each of these surves is horizontal

Answer 1

a) y = xlogx
b) y = (1/x) + logx
c) y = (x^2)log(1/x)

Answer 2

for a horizontal tangent dy/dx=0 so differentiate each thing and equate to 0 so for a) logx+1=0 logx=-1 x=1/e. The rest are doable.

Answer 3

logx=-1 x=1/e. could you please explain how you got this

Answer 4

well taking anti log on both sides.

Answer 5

i assumed(Mostly it is so unless specified otherwise) logs in calculus are to the base e not 10 hence if log a to base e = k then a=e^k in this case k=-1 e^-1 is 1/e

Answer 6

ahh i see thanks