Question

How do you find the equation of the tangent line to the graph `f(x)=ln((e^x+e^-x)/2)` through point (0,0)?

Answer 1

The x-axis, `larr y = 0 rarr`. See the tangent-inclusive Socratic graph.

Explanation:

`f=lncoshx and f'=1/coshx(coshx)'=sinhx/coshx=0/1=0`, at `x = 0.`

So, the equation to the tangent at (0, 0) is

`y-0 = 0(x-0)=0`

graph{(e^y-(e^x+e^(-x))/2)y=0 [-10, 10, -5, 5]}