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(C) Given the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$.First,we find the critical points by setting the derivative $f'(x) = 0$.$f'(x) =

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$4$

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(C) Given the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$.First,we find the critical points by setting the derivative $f'(x) = 0$.$f'(x) = 2x e^{2x} + x^2 (2 e^{2x}) = 2x e^{2x} (1 + x)$.Setting $f'(x) = 0$,we get $x = 0$ or $x = -1$.Now,we evaluate the function at the critical points and the endpoints of the interval $[-2, 2]$:$f(-2) = (-2)^2 e^{2(-2)} = 4 e^{-4}$.$f(-1) = (-1)^2 e^{2(-1)} = 1 e^{-2} = e^{-2}$.$f(0) = (0)^2 e^{2(0)} = 0$.$f(2) = (2)^2 e^{2(2)} = 4 e^4$.Comparing these values,the global maximum $p = 4 e^4$ and the global minimum $q = 0$.Finally,we calculate $p e^{-4} + q e^4 = (4 e^4) e^{-4} + (0) e^4 = 4 e^0 + 0 = 4(1) = 4$.

If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$,then $p e^{-4} + q e^4 =$

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