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

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Answer 1 · 2017-02-01T01:48:17

The x-axis, $\leftarrow y = 0 \to$ $larr y = 0 rarr$ . See the tangent-inclusive Socratic graph.

$f=lncoshx and f'=1/coshx(coshx)'=sinhx/coshx=0/1=0$ , 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]}

How do you find the equation of the tangent line to the graph f(x)=ln((e^x+e^-x)/2)f(x)=ln(ex+e−x2)f(x)=ln((e^x+e^-x)/2) through point (0,0)?