Find the absolute extrema of the function. (Round your answer to three decimal places.)
f(x) = xe-x2 on [0,2]
Absolute maximum value:
at x =
Absolute minimum value:
at x =
f = xe^-x^2
f' = (1-2x^2) e^-x^2
so, where is f' = 0?
To find the absolute extrema of the function f(x) = xe^{-x^2} on the interval [0,2], we need to determine the maximum and minimum values of the function within this interval.
To find the critical points, we need to find where the derivative equals zero or is undefined.
Step 1: Find the derivative of the function.
f'(x) = e^{-x^2}(1-2x^2)
Step 2: Set the derivative equal to zero and solve for x.
e^{-x^2}(1-2x^2) = 0
Since the exponential term e^{-x^2} is never zero, the critical points occur when (1-2x^2) = 0.
Solving for x gives us:
1 - 2x^2 = 0
2x^2 = 1
x^2 = 1/2
x = ± (1/√2)
However, we need to check if these critical points lie within the interval [0,2].
Since the interval is [0,2], we only need to check if x = 1/√2 lies within this range.
Since 1/√2 is greater than 0 and less than 2, it is within the interval [0,2].
Step 3: Evaluate the function at the critical points and the endpoints of the interval.
f(0) = 0e^{-0^2} = 0
f(2) = 2e^{-2^2} = 2e^{-4}
f(1/√2) = (1/√2)e^{-(1/√2)^2}
= (1/√2)e^{-1/2}
= (1/√2)(1/e^{1/2})
= 1/(e^{1/2√2})
Step 4: Compare the function values to determine the absolute extrema.
The maximum value will be the largest value among f(0), f(2), and f(1/√2).
The minimum value will be the smallest value among f(0), f(2), and f(1/√2).
f(0) = 0
f(2) = 2e^{-4}
f(1/√2) = 1/(e^{1/2√2})
Comparing these values, we can see that the absolute maximum value occurs at:
at x = 2
And the absolute minimum value occurs at:
at x = 0
To find the absolute extrema of the function f(x) = xe^(-x^2) on the interval [0, 2], we need to find the highest and lowest points on that interval.
Step 1: Find the critical points of the function within the interval [0, 2]. Critical points occur when the derivative of the function is equal to zero or undefined.
To find the derivative of f(x), we can use the product rule:
f'(x) = (1)(e^(-x^2)) + (x)(-2x)(e^(-x^2))
f'(x) = e^(-x^2) - 2x^2e^(-x^2)
f'(x) = e^(-x^2)(1 - 2x^2)
To find the critical points, we set f'(x) = 0 and solve for x:
e^(-x^2)(1 - 2x^2) = 0
Since e^(-x^2) is never zero, we have:
1 - 2x^2 = 0
2x^2 = 1
x^2 = 1/2
x = ±√(1/2)
x = ±(1/√2) = ±(1/√2) * (√2/√2) = ±(√2/2)
So, the critical points within the interval [0, 2] are x = (√2/2) and x = -(√2/2).
Step 2: Evaluate the function at the critical points and at the endpoints of the interval [0, 2].
f(0) = (0)(e^(-(0^2))) = 0
f(2) = (2)(e^(-(2^2))) = 2e^(-4)
f(√2/2) = (√2/2)(e^(-((√2/2)^2))) = (√2/2)e^(-1/2) ≈ 0.347
f(-√2/2) = (-√2/2)(e^(-((-√2/2)^2))) = (-√2/2)e^(-1/2) ≈ -0.347
Step 3: Compare the values obtained. The absolute maximum value will be the highest value found, and the absolute minimum value will be the lowest value found.
- Absolute maximum value: The highest value is f(2) = 2e^(-4) ≈ 0.036.
- Absolute minimum value: The lowest value is f(-√2/2) ≈ -0.347.
Therefore, the absolute extrema of the function f(x) = xe^(-x^2) on the interval [0, 2] are:
- Absolute maximum value: 0.036 (at x = 2)
- Absolute minimum value: -0.347 (at x = -√2/2)