Final answer:
The function f(x) = x²e⁻¹x has a local maximum at (1, 1/e) because the derivative changes sign from positive to negative. There is no absolute minimum since the function approaches zero but never becomes negative. The question's statement that f(0) = 0.25 is incorrect; f(0) is actually 0. The correct answer is b) Local Max: (1, 1/e), Absolute Min: None.
Step-by-step explanation:
To find the local maximum and absolute minimum of the function f(x) = x²e⁻¹x, one must first take the derivative of the function to determine the critical points. The derivative, f'(x), can be found using the product rule:
f'(x) = 2xe⁻¹x + x²(-e⁻¹x)
Simplifying this, we get:
f'(x) = (2 - x)x²e⁻¹x
To find the critical points, set f'(x) to zero:
0 = (2 - x)x²e⁻¹x
This equation equals zero when x = 0 or x = 2. To determine whether these points are maxima, minima, or inflection points, we can use the first and second derivative tests.
We know that the exponential function is always positive, so we just need to analyze the sign of (2 - x)x². At x = 0, there is a local maximum because f'(x) changes from positive to negative as x increases. At x = 1, the first derivative is zero and f''(1) will be negative, indicating a local maximum.
The function value at x = 1 is f(1) = 1/e, and since e is approximately 2.718, this is less than one. The question incorrectly states that f(0) = 0.25, but in fact, f(0) = 0 since x² will be zero at x = 0. Therefore, the local maximum is at (1, 1/e), and there is no absolute minimum since the function approaches zero as x goes to infinity, and it is never negative.
Thus, the correct choice is:
Local Max: (1, 1/e), Absolute Min: None