The point for the curve, `y=xe^(x)`,

Options

A. `x=-1` is minima
B. `x=0` is minima
C. `x=-1` is maxima
D. `x=0` is maxima

Explanation

To determine the nature of the critical points for the curve \( y = x e^x \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = x e^x \) using the product rule. The product rule states that if you have two functions \( u \) and \( v \), then the derivative \( \frac{d}{dx}(uv) = u'v + uv' \).

Here, let:
- \( u = x \) and \( v = e^x \)

Thus, we have:
- \( u' = 1 \)
- \( v' = e^x \)

Now applying the product rule:
\[ \frac{dy}{dx} = u'v + uv' = 1 \cdot e^x + x \cdot e^x = e^x + x e^x = e^x (1 + x) \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:
\[ e^x (1 + x) = 0 \]

Since \( e^x \) is never zero, we can simplify this to:
\[ 1 + x = 0 \implies x = -1 \]

### Step 3: Determine the second derivative
To determine whether this critical point is a maximum or minimum, we need to find the second derivative \( \frac{d^2y}{dx^2} \).

We differentiate \( \frac{dy}{dx} = e^x (1 + x) \) again using the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x) (1 + x) + e^x \frac{d}{dx}(1 + x) \]
Calculating each term:
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( 1 + x \) is \( 1 \).

Thus, we have:
\[ \frac{d^2y}{dx^2} = e^x (1 + x) + e^x \cdot 1 = e^x (1 + x + 1) = e^x (2 + x) \]

### Step 4: Evaluate the second derivative at the critical point
Now we evaluate the second derivative at the critical point \( x = -1 \):
\[ \frac{d^2y}{dx^2} \bigg|_{x = -1} = e^{-1} (2 - 1) = e^{-1} \cdot 1 = \frac{1}{e} \]

Since \( \frac{1}{e} > 0 \), this indicates that the function is concave up at \( x = -1 \), which means that \( x = -1 \) is a local minimum.

### Conclusion
Thus, the point \( x = -1 \) is a minimum for the curve \( y = x e^x \).