For what values of x does f(x)=(secx)/(1+tanx) have a horizontal tangent?

Answer 1

The equation of the curve is f(x) =secx/(1+tanx)

Simply the curve equation.

f(x) =(1/cosx)/(1+(sinx/cosx))

      =1/(sinx + cosx)

Find the derivative of the curve:

Quotient rule of differentiation:

d/dx(u/v)=(vu' - uv')/ v²

u= 1    u' = 0

v = sinx + cosx  v'=cosx - sinx

f'(x) =((sinx + cosx) d/dx (1) -(1) d/dx(sinx + cosx)) /(sinx + cosx)²

       =(0-(cosx - sinx)) /(sinx + cosx)²

       =(sinx - cosx) /(sinx + cosx)²

Horizontal tangent slope is zero .

m =f'(x)=0

(sinx - cosx) /(sinx + cosx)²

sinx - cosx=0

sinx = cosx

tan x = 1

tan x = tan π/4

General solution is x = nπ +π/4  n Є z