y=x*e-x

It looks like you're working with the function \( y = xe^{-x} \). If you're preparing for exams, you might want to analyze this function further! Here are some common aspects you might explore:

1. Domain and Range:

The function \( y = xe^{-x} \) is defined for all \( x \in \mathbb{R} \). As for the range, since \( e^{-x} \) approaches 0 as \( x \) goes to infinity and negative infinity, the range is \( (-\infty, \infty) \).

2. Intercepts:

• y-intercept: Set \( x = 0 \): \[ y(0) = 0 \cdot e^0 = 0 \] So, the y-intercept is \( (0, 0) \).

• x-intercepts: Set \( y = 0 \): \[ xe^{-x} = 0 \implies x = 0 \] Thus, the x-intercept is also \( (0, 0) \).

3. Finding the Derivative:

To find critical points and analyze the function, we can find the derivative using the product rule:

\[ y' = \frac{d}{dx}(xe^{-x}) = e^{-x} + x\frac{d}{dx}(e^{-x}) \] Using the chain rule, \( \frac{d}{dx}(e^{-x}) = -e^{-x} \):

\[ y' = e^{-x} + x(-e^{-x}) = e^{-x} - xe^{-x} = e^{-x}(1 - x) \]

4. Finding Critical Points:

Set the derivative \( y' = 0 \): \[ e^{-x}(1 - x) = 0 \] Since \( e^{-x} \neq 0 \) for any finite \( x \), we have: \[ 1 - x = 0 \implies x = 1 \]

5. Second Derivative Test:

To determine whether this critical point is a maximum or minimum, we can compute the second derivative:

\[ y'' = \frac{d}{dx}(e^{-x}(1 - x)) \]

Using the product rule: \[ y'' = e^{-x}(1 - x)(-1) + e^{-x}(-1) = -e^{-x}(1 - x) - e^{-x} \] \[ y'' = -e^{-x}(2 - x) \]

Evaluate \( y'' \) at \( x = 1 \): \[ y''(1) = -e^{-1}(2 - 1) = -e^{-1} < 0 \]

Thus, \( x = 1 \) is a local maximum.

6. Value at Critical Point:

Evaluate \( y \) at \( x = 1 \): \[ y(1) = 1 \cdot e^{-1} = \frac{1}{e} \]

7. Behavior at Infinity:

As \( x \to \infty \), \( y = xe^{-x} \to 0 \) since the exponential decay dominates the linear growth. As \( x \to -\infty \), \( y \to 0 \).

Summary:

• The function has a local maximum at \( (1, \frac{1}{e}) \).
• The function is increasing on \( (-\infty, 1) \) and decreasing on \( (1, \infty) \).
• The x- and y-intercept is at the origin \( (0, 0) \).

If you have specific questions or need to explore other concepts related to this function, feel free to ask!