Math  /  Calculus

QuestionConsider the function given by f(x)=x2e2xf(x)=x^{2} e^{-2 x}
Determine the absolute maximum value and absolute minimum value of ff over the interval [1/2,3][-1 / 2,3]. FORMATTING: Give your answer with an accuracy of at least 3 decimal places.
Minimum value == \square Maximum value == Number

Studdy Solution
Compare the values:
1. f(1/2)=14e0.679 f(-1/2) = \frac{1}{4}e \approx 0.679
2. f(0)=0 f(0) = 0
3. f(1)=e20.135 f(1) = e^{-2} \approx 0.135
4. f(3)=9e60.002 f(3) = 9e^{-6} \approx 0.002

The absolute minimum value is f(0)=0 f(0) = 0 .
The absolute maximum value is f(1/2)0.679 f(-1/2) \approx 0.679 .
Minimum value =0 = 0 Maximum value =0.679 = 0.679

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