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Let $b \in \mathbb{R}$ and $a \in \mathbb{R}\setminus\{0\}$, I'd like to prove that $$\tanh(x) = ax + b$$ has at least one, but at most three solutions. That the equation has always at least one solution follows by the intermediate value theorem (i.e., both sides are real continuous functions and $\tanh(x)$ is bounded, while $ax + b$ is not).

Yet, I lack a concrete approach to prove that the equation has at most three solutions (even though this is easily apparent from a graphical observation).

rico
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  • Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Jan 16 '23 at 09:39
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    @runway44 This question uses the hyperbolic tangent, not the regular one. – Arthur Jan 16 '23 at 09:44
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    You said you did "graphical observations". Well, what did your learn about different values for $a, b$? If you found something useful, you can definitely work backwards to obtain a concrete proof. – Dstarred Jan 16 '23 at 09:50
  • If we could prove that the second derivate of tanh(x) has only one root, namely $0$, we could use Rolle's theorem. – Peter Jan 16 '23 at 09:57
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    Other info; see that $\max(\tanh{x})' = 1$ at $x = 0$ and $\lim_{x \to \pm \infty}{(\tanh{x})'} \to 0$. What does this indicate about particular values of $a$? – Dstarred Jan 16 '23 at 09:57
  • Hi. There is a similar question here: https://math.stackexchange.com/questions/4029249/proving-that-there-are-3-solutions-at-most-fx-ex-ax2bxc/4617104#4617104 – Bob Dobbs Jan 16 '23 at 10:36

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Let $g(x)=\tanh (x)-ax-b$ , Since $(g(+\infty),g(-\infty))=(+\infty,-\infty)$ or $(-\infty,+\infty)$, by intermediate value theorem, $g(x)=0$ has at least one solution. $g'(x)=\operatorname{sech}^2(x)-a$, so $g'(x)=0$ has at most two solutions, by Rolle's theorem, $g(x)=0$ has at most three solutions.

Gary
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