Question 1042538
The amplitude is normally {{{ 1 }}}, so you
need a factor of {{{ 2 }}}
The midline is normally {{{ 0 }}}, so you need
to add {{{ 1 }}} to the function
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So far, it looks like
{{{ f(x) = 2*cos( k*x ) + 1 }}}
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You ar told that:
when {{{ x = pi/3 }}} then the cosine has gone 
through an entire period, so it must be true:
{{{ k*( pi/3 ) = 2*pi }}}
{{{ k = 6 }}}
So, the function must be:
{{{ f(x) = 2*cos( 6x ) + 1 }}}
Here's a plot of 2 periods in both directions from 
{{{ x = 0 }}}. It is plotted from {{{ x = -2*pi/3 }}} to
{{{ x = 2*pi/3 }}}
{{{ graph( 800, 300, -(2*pi/3), 2*pi/3, -1, 3, 2*cos(6x) + 1  ) }}}
( note that {{{ f(pi/3) = 3 }}} as it should )