y = e^tanx - 2
To find at which point it crosses x axis we state that y= 0
e^tanx - 2 = 0
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061
to find slope at that point first we need to find first derivative of funtion y.
y' = (e^tanx)*1/cos^2(x)
now we express x = 0.6061 in y' and we get:
y' = k = 2,9599
