Show that $\sinh x$ is strictly increasing Can someone help me with this analysis question:
I have to show that $\sinh(x)$ is strictly increasing on all of $\Bbb R$, and that $\cosh(x)$ is strictly increasing on the interval: $[0,\infty)$ and strictly decreasing on the interval: $(-\infty, 0]$. 
I'm not allowed to differentiate or anything like that. I think I have to do it with inequalities.  
I've been giving a hint to use the trigonometric addition formulas.
 A: exp(x) is increasing.
exp(-x) is decreasing, so -exp(-x) is increasing.
The sum of two increasing functions is increasing.
A: It's easy to see that $\sinh(-x)=-\sinh(x)$ and $\cosh(-x)=\cosh(x)$, so it suffices to show that each function is increasing on $[0,\infty)$.  Their increasing/decreasing behaviors on $(-\infty,0]$ follow from their odd/even symmetries.
For $\sinh(x)$, this is easy, provided you know that $e^x$ is an increasing function:  That makes $e^{-x}$ a decreasing function, which makes $-e^{-x}$ an increasing function, and that makes their sum increasing. (This argument actually works for all $x$, not just for $x\ge0$.  It's the next paragraph that uses the restriction.)
For $\cosh(x)$, it's a little trickier. since you are summing a decreasing function to an increasing function.  But here we're aided by that fact that $\sinh(x)\ge0$ for all $x\in[0,\infty)$, so that $\sinh^2(x)$, being the product of non-negative increasing functions, is also increasing on $[0,\infty)$, which means $\cosh(x)=\sqrt{\sinh^2(x)+1}$ in increasing (being a composition with the increasing square root function).
Added later:  Here is a second approach that may be more along the lines of what the OP is looking for.  Again, we'll restrict to $x\ge0$ (using odd/even symmetry for negative $x$), and we'll take as given that $\sinh(x)$ and $\cosh(x)$ are both non-negative for $x\ge0$, from which the hyperbolic identity $\cosh^2(x)-\sinh^2(x)=1$ implies $\cosh(x)=\sqrt{1+\sinh^2(x)}\ge1$.
From the addition formulae
$$\sinh(x+u)=\sinh(x)\cosh(u)+\cosh(x)\sinh(u)$$
and
$$\cosh(x+u)=\cosh(x)\cosh(u)+\sinh(x)\sinh(u)$$
we have, for $x,u\ge0$,
$$\sinh(x+u)\ge\sinh(x)\cdot1+0=\sinh(x)$$
and
$$\cosh(x+u)\ge\cosh(x)\cdot1+0=\cosh(x)$$
so that $\sinh(x)$ and $\cosh(x)$ are non-decreasing on $[0,\infty)$.  If we know (or prove) that $\sinh(u)$ is positive for $u\gt0$, then $\cosh(u)\gt1$, and we get $\sinh(x+u)\gt\sinh(x)$ and $\cosh(x+u)\gt\cosh(x)$, which means $\sinh(x)$ and $\cosh(x)$ are strictly increasing on $[0,\infty)$.
A: if $f(x)=\frac{e^x-e^{-x}}{2}$ then we have $$f'(x)=\frac{e^x+e^{-x}}{2}>0$$ and our function i strictly monotonously increasing.
A: $f(x)=\exp(x)$ and $g(x)=- \exp(-x)$ are both increasing functions.
Therefore, so is $\sinh(x) = \frac12(f(x)+g(x))$.
A: $$\sinh x=\sum_{k\ge 0}\frac{x^{2k+1}}{(2k+1)!},$$
and each $x^{2k+1}$ is an increasing function.
Similar argument for $$\cosh x=\sum_{k\ge 0}\frac{x^{2k}}{(2k)!}.$$
