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 find the points of the curve given by the equation \( y = x e^x \), we need to determine the critical points by finding the first derivative, setting it to zero, and analyzing the second derivative to classify these critical points.

### Step 1: Find the first derivative
The function is given as:
\[ y = x e^x \]

To find the first derivative \( \frac{dy}{dx} \), we will use the product rule, which states that if \( y = u \cdot v \), then:
\[ \frac{dy}{dx} = u'v + uv' \]

Here, let:
- \( u = x \) (so \( u' = 1 \))
- \( v = e^x \) (so \( v' = e^x \))

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

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

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

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

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

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

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

### Conclusion
Thus, the point for the curve \( y = x e^x \) where there is a minimum is at:
\[ x = -1 \]