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= the Absolute Maximum of the Function f(x)=x42x2+3f ( x ) = x ^ { 4 } - 2 x ^ { 2 } + 3

Question 45

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= The absolute maximum of the function f(x)=x42x2+3f ( x ) = x ^ { 4 } - 2 x ^ { 2 } + 3 on the interval -1 \le x \le 2 is 11.

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