If the function f(x) = xe -x , x ∈ R attains its maximum value β at x = α then (α, β) =
\((2,\frac{1}{e})\)
\((1, \frac{1}{e})\)
\((2,\frac{-1}{e})\)
\((\frac{1}{e}, 1)\)
To solve the problem, we need to find the maximum value $\beta$ of the function $f(x) = x e^{-x}$ and the corresponding $x = \alpha$.
1. Find the Critical Points:
Compute the first derivative:
$f'(x) = e^{-x} + x(-e^{-x}) = e^{-x} - x e^{-x} = e^{-x}(1 - x)$.
Set $f'(x) = 0$:
$e^{-x}(1 - x) = 0$. Since $e^{-x} \neq 0$, we have $1 - x = 0$, so $x = 1$.
2. Determine the Nature of the Critical Point:
Compute the second derivative:
$f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 1 - 1) = e^{-x}(x - 2)$.
Evaluate at $x = 1$:
$f''(1) = e^{-1}(1 - 2) = -\frac{1}{e} < 0$, indicating a maximum.
3. Calculate the Maximum Value:
At $x = 1$, the function value is:
$f(1) = 1 \cdot e^{-1} = \frac{1}{e}$.
Thus, $\alpha = 1$ and $\beta = \frac{1}{e}$, so $(\alpha, \beta) = \left(1, \frac{1}{e}\right)$.
Final Answer:
The maximum occurs at $\left(1, \frac{1}{e}\right)$.
Match the following:
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Integral calculus helps to resolve two major types of problems: