My book asked me to prove that the function $\sinh x$ is odd, but in order to be odd I must be sure that the domain of it symmetric about the origin, how can I be sure from this?
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The usual definition of $\sinh x$ is $$\sinh x=\frac{1}{2} (e^x-e^{-x})$$ Now use the definition of an odd function $f(x)=-f(-x)$ to prove that $\sinh x$ is odd – Yuriy S Oct 23 '18 at 08:59
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"My book asked me to prove that the function $\sinh x $ is odd" - there's no talk about domains here, you just need to prove it's odd according to the definition of an odd function – Yuriy S Oct 23 '18 at 09:34
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2@YuriyS: Yes, there is a condition on the domain, but since the domain in this case is the whole real line, it is trivially satisfied. – Hans Lundmark Oct 23 '18 at 09:45
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@HansLundmark, thank you for the correction – Yuriy S Oct 23 '18 at 09:46
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$$f(x)=\sinh x:=\frac{e^x-e^{-x}}2=\color{red}-\frac{e^{-x}-e^x}2=-\sinh(-x)=-f(-x)$$
DonAntonio
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1@hopefully Fair enough: what is your definition of "symmetric domain"? – DonAntonio Oct 23 '18 at 09:09
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@hopefully That doesn't say much to me...By definition, a function $;g;$ with a definition domain symmetric wrt the origin (meaning:a domain of the form $;(-a,a);,;;a\in\Bbb R;$ ) is odd if $;g(-x)=-g(x);$ ... – DonAntonio Oct 23 '18 at 09:36
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@hopefully...Well, read again carefully my answer...Do you know the exponential function $;e^x;$ ? Do you know it is defined everywhere? – DonAntonio Oct 23 '18 at 09:39
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