Asymptotes to tanh I am new to this site. I have just started reading hyperbolic functions in calculus. I wanted to know if we could derive the shape of y=tanh x using derivative tests ( if yes, how?). By shape I mean, It has two horizontal asymptotes at x=1, x=−1. it is inverted u in shape when x belongs to R+ It is U shaped when x belongs to R−

rkus2zg0

rkus2zg0

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2022-08-20

Asymptotes to tanh
I am new to this site. I have just started reading hyperbolic functions in calculus. I wanted to know if we could derive the shape of y = tanh x using derivative tests ( if yes, how?). By shape I mean,
It has two horizontal asymptotes at x=1, x=−1.
it is inverted u in shape when x belongs to R +
It is U shaped when x belongs to R

Answer & Explanation

Bryce Obrien

Bryce Obrien

Beginner2022-08-21Added 8 answers

The asymptotes y = ± 1 are immediate from tanh x = e x + e x e x e x by noticing that the first terms dominate towards + giving 1 + 0 1 0 = + 1 and the second terms dominate towards giving 0 + 1 0 1 = 1. I'm not sure you can use a derivative test to establish this, though.
For the others, just note that
( tanh x ) = 1 cosh 2 x
( tanh x ) = 2 sinh x cosh 3 x
Recall that coshx is always positive, and that sinhx has the sign of x. Thus:
-- The first derivative is always positive (so the function is strictly increasing everywhere).
-- The second derivative is positive for x<0 (so the function is concave up there), and is negative for x>0 (so the function is concave down there).
That's pretty much everything you mentioned, right?

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