proof $|\sinh(x)|\leq|3x|$ for $ -\frac{1}{2}I am supposed to prove that $|\sinh(x)|\leq3|x|$ for $|x|<\frac{1}{2}$. 
I know I am supposed to use $|e^x-1|\leq3|x|$ for $|x|<\frac{1}{2}$. 
I am completely stuck, and I don't know how to approach this, so any help is greatly appreciated!

Since $|\sinh{(-x)}|=|\sinh{x}|$, it's enough to prove that $0\leq\sinh{x}\leq3x$ for $0\leq x<\frac{1}{2}$ or
$$0\leq e^x-e^{-x}\leq6x.$$
The left inequality is true because $e^x-e^{-x}=\frac{(e^x-1)(e^x+1)}{e^x}\geq\frac{(e^0-1)(e^x+1)}{e^x}=0.$
We'll prove the right inequality.
Let $f(x)=6x-e^x+e^{-x}$.
Hence, $$f'(x)=6-e^x-e^{-x}=-\frac{e^{2x}-6e^x+1}{e^x}=\frac{(3+2\sqrt2-e^x)(e^x-(3-2\sqrt2))}{e^x}\geq$$
$$\geq\frac{(3+2\sqrt2-e^{\frac{1}{2}})(e^0-(3-2\sqrt2))}{e^x}>0,$$
which says that $f$ is an increasing function on $[0,\frac{1}{2}]$.
Thus, $f(x)\geq f(0)=0$ and we are done!
