Question #2be8a

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Answer 1 · 2016-07-20T08:53:10

$f(x) = cosh x = (e^x + e^-x)/(2)$

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

$f''(x) = 1/2( e^x + e^-x ) = cosh x$

and it's pretty simple from there

$f'(x) = sinh x + C$

$f'(0) = 0 implies 0 = sinh (0) + C implies C = 0$

$f(x) = cosh x + C$

$f(0) = 1 implies 1 = cosh 0 + C implies C = 0$

$therefore f(x) = cosh x = (e^x + e^-x)/(2)$