Question 1062776
{{{ f(x) = (x-2)^2 + 4 }}}
Take the derivative to get:
f'(x) = {{{  2(x-2) }}}
f'(x) = {{{ 0 = 2x-4 }}}  —> {{{ x = 2 }}}

x=2 is a critical point (it could be a local minimum, local maximum, or just a flat spot in the function)

Method 1: 
f(2) = 4
Also plug in values near x=2, on either side of 2, to see what f(x) does
f(3) = 5
f(1) = 5
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Ans:  {{{highlight( x=2 )}}} is where the function's minimum occurs
(If you are ever given a domain [a,b] of x values, be sure to also compare f(a) and f(b))
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Method 2:
f'(x) = {{{ 2(x-2) = 2x-4 }}}
Take 2nd derivative and see if function is concave up or down.
f''(x) = {{{ 2 }}}  —>   concave up, therefore {{{ highlight(x=2)}}} is where f(x) is minimum