Show that $\sinh(\mathbb{R})=\mathbb{R}$
I know that $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$ but I can't see how inputting the set of real numbers gets the real numbers back as $e^x \gt 0$
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Note that $-e^{-x}<0$ and $-e^{-x} \to -\infty$ as $x\to-\infty,$ while $e^x\to0$ as $x\to-\infty.$
This function
is everywhere increasing, as demonstrated by the fact that its derivative is everywhere positive, and
is everywhere continuous, and
approaches $\pm\infty$ as $x\to\pm\infty$ respectively.
Thus the intermediate value theorem can be used to show that the image of this function is all of $\mathbb R.$
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So because it approaches infinity from both sides then it also takes any value in between? From the intermediate value theorem – Joanna M Feb 08 '18 at 03:03
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@JoannaM: Except it doesn't approach infinity from both sides (how would that be possible?!?), it approaches “both infinities”. $+\infty$ and $-\infty$. – Hans Lundmark Feb 08 '18 at 06:44
...as e^x > 0The difference of two positive numbers can be negative. Check out the limits at $\pm \infty$. – dxiv Feb 08 '18 at 02:47