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Math  /  Calculus

QuestionConsider the function given by f(x)=x2e−2xf(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

STEP 1

1. The function f(x)=x2e−2x f(x) = x^2 e^{-2x} is continuous over the interval [−1/2,3][-1/2, 3].
2. To find absolute extrema (maximum and minimum), we need to evaluate the function at critical points and endpoints.
3. Critical points occur where the derivative is zero or undefined.

STEP 2

1. Find the derivative of f(x) f(x) .
2. Determine the critical points by setting the derivative equal to zero.
3. Evaluate f(x) f(x) at the critical points and endpoints of the interval.
4. Compare these values to determine the absolute maximum and minimum.

STEP 3

Find the derivative of f(x)=x2e−2x f(x) = x^2 e^{-2x} .
Using the product rule, where u=x2 u = x^2 and v=e−2x v = e^{-2x} , the derivative is:
f′(x)=(x2)′e−2x+x2(e−2x)′ f'(x) = (x^2)' e^{-2x} + x^2 (e^{-2x})'
f′(x)=2xe−2x+x2(−2e−2x) f'(x) = 2x e^{-2x} + x^2 (-2e^{-2x})
f′(x)=2xe−2x−2x2e−2x f'(x) = 2x e^{-2x} - 2x^2 e^{-2x}
f′(x)=2xe−2x(1−x) f'(x) = 2xe^{-2x} (1 - x)

STEP 4

Determine the critical points by setting the derivative equal to zero:
2xe−2x(1−x)=0 2xe^{-2x} (1 - x) = 0
This gives us:
1. 2x=0 2x = 0 which implies x=0 x = 0
2. 1−x=0 1 - x = 0 which implies x=1 x = 1

STEP 5

Evaluate f(x) f(x) at the critical points and endpoints of the interval [−1/2,3][-1/2, 3].
1. Evaluate at x=−1/2 x = -1/2 :
f(−1/2)=(−12)2e−2(−1/2)=14e1=14e f(-1/2) = \left(-\frac{1}{2}\right)^2 e^{-2(-1/2)} = \frac{1}{4} e^1 = \frac{1}{4}e
2. Evaluate at x=0 x = 0 :
f(0)=02e−2(0)=0 f(0) = 0^2 e^{-2(0)} = 0
3. Evaluate at x=1 x = 1 :
f(1)=12e−2(1)=e−2 f(1) = 1^2 e^{-2(1)} = e^{-2}
4. Evaluate at x=3 x = 3 :
f(3)=32e−2(3)=9e−6 f(3) = 3^2 e^{-2(3)} = 9e^{-6}

STEP 6

Compare the values:
1. f(−1/2)=14e≈0.679 f(-1/2) = \frac{1}{4}e \approx 0.679
2. f(0)=0 f(0) = 0
3. f(1)=e−2≈0.135 f(1) = e^{-2} \approx 0.135
4. f(3)=9e−6≈0.002 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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