# Question #2be8a

Jul 20, 2016

$f \left(x\right) = \cosh x = \frac{{e}^{x} + {e}^{-} x}{2}$

#### Explanation:

well, you can do it super quickly by recognising it is hyperbolic.

$f ' ' \left(x\right) = \frac{1}{2} \left({e}^{x} + {e}^{-} x\right) = \cosh x$

and it's pretty simple from there

$f ' \left(x\right) = \sinh x + C$

$f ' \left(0\right) = 0 \implies 0 = \sinh \left(0\right) + C \implies C = 0$

$f \left(x\right) = \cosh x + C$

$f \left(0\right) = 1 \implies 1 = \cosh 0 + C \implies C = 0$

$\therefore f \left(x\right) = \cosh x = \frac{{e}^{x} + {e}^{-} x}{2}$