The curve `y=xe^(x)` has minimum value equal to:

Options

A. `-1/e`
B. `1/e`
C. `-e`
D. e

Answer

To find the minimum value of the curve \( y = x e^x \), we will follow these steps:

### Step 1: Find the first derivative
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 \( \frac{d}{dx}(uv) = u'v + uv' \).

Let:
- \( u = x \) and \( v = e^x \)

Now, we differentiate:
- \( u' = 1 \)
- \( v' = e^x \)

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

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

Since \( e^x \) is never zero, we can focus on the term \( x + 1 = 0 \):
\[ x + 1 = 0 \implies x = -1 \]

### Step 3: Determine if it is a minimum or maximum using the second derivative
Next, we need to find the second derivative to determine whether this critical point is a minimum or maximum.

We differentiate \( \frac{dy}{dx} = e^x (x + 1) \) again:
Using the product rule again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x) (x + 1) + e^x \frac{d}{dx}(x + 1) \]
Calculating each part:
- \( \frac{d}{dx}(e^x) = e^x \)
- \( \frac{d}{dx}(x + 1) = 1 \)

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

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

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

### Step 5: Calculate the minimum value
Finally, we find the minimum value of the function by substituting \( x = -1 \) back into the original function:
\[ y = (-1) e^{-1} = -\frac{1}{e} \]

### Conclusion
The minimum value of the curve \( y = x e^x \) is:
\[ -\frac{1}{e} \]