Show that $\sinh x \ge 0$ for all $x\ge 0$

Question

Can someone help me show what's written in the title? I know that $\cosh^2x - \sinh^2x=1$

Note: Not allowed to use derivatives to solve it

Answer 1

Remember that $\sinh(x)=\frac{e^x-e^{-x}}2$. Now try to calculate $\sinh(0)$ and analyze its derivative for $x\ge 0$, and see if you can get to some conclusion.

Answer 2

note that $$(\sinh(x))'=\cosh(x)\geq 0$$ for all real $x$ and so also for $x\geq 0$

Answer 3

For $x\geq 0$, $e^{x}-e^{-x}\geq 0$ if and only if $e^{x}\geq e^{-x}$ if and only if $e^{2x}\geq 1$, and this is true for $x\geq 0$.

Answer 4

Hint

Use $ \sinh(x)=\dfrac {e^x-e^{-x}}{2} $

$$\frac {e^x-e^{-x}}{2} \geq 0$$ $$\frac {e^{2x}-1}{2e^x} \geq 0$$ $$ e^{2x} \geq 1=e^0$$ $$x \geq 0$$

For $x \geq 0 , f(x)=e^x $ increase and equal 1 for $x=0$

Answer 5

$$\sinh(x)=\frac{e^x-e^{-x}}{2}=$$ $$\frac{\sum\limits_{n\geq 0} \frac{x^n}{n!}-\sum\limits_{n\geq 0} \frac{(-x)^n}{n!}}{2}=$$ $$\frac{\sum\limits_{\text{odd } n} \frac{x^{n}}{n!}+\sum\limits_{\text{even } n} \frac{x^{n}}{n!}+\sum\limits_{\text{odd } n} \frac{x^{n}}{n!}-\sum\limits_{\text{even } n} \frac{x^{n}}{n!}}{2}=$$ $$\frac{2\sum\limits_{\text{odd } n} \frac{x^{n}}{n!}}{2}=\sum\limits_{\text{odd } n} \frac{x^{n}}{n!} >0\text{ for } x>0$$ And $\sinh(0)=0$