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Response Details:
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Set F'(x)=0 and solve for x to find critical numbers

Since x=-1 is outside of [0,2], so we choose x=1

At x=1, F '(x) changes sign from positive to negative, so by the First Derivative Test, F(x) has local maximum (and absolute maximum) at (1,1/2) and F(x) has absolute minimum at (0,0)

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