For the curve `y=x e^x` , the point

Options

A. x=-1 is a point of minimum
B. x=0 is a point of minimum
C. x=-1 is a point of maximum
D. x=0 is a point of maximum

Answer

To determine whether the points \( x = -1 \) or \( x = 0 \) are points of maximum or minimum for the curve \( y = x e^x \), we will follow these steps:

### Step 1: Write the function
We start with the function:
\[ y = x e^x \]

### Step 2: Find the first derivative
To find the critical points, we need to calculate the first derivative \( \frac{dy}{dx} \). We will use the product rule, which states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \).

Let \( u = x \) and \( v = e^x \).

Calculating the derivatives:
- \( u' = 1 \)
- \( v' = e^x \)

Now applying the product rule:
\[ \frac{dy}{dx} = u'v + uv' = 1 \cdot e^x + x \cdot e^x = e^x + x e^x \]
Thus, we can factor this as:
\[ \frac{dy}{dx} = e^x (x + 1) \]

### Step 3: 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 simplify this to:
\[ x + 1 = 0 \implies x = -1 \]

### Step 4: Find the second derivative
Next, we need to find the second derivative \( \frac{d^2y}{dx^2} \) to determine the nature of the critical point. We differentiate \( \frac{dy}{dx} = e^x (x + 1) \) again.

Using the product rule again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x) \cdot (x + 1) + e^x \cdot \frac{d}{dx}(x + 1) \]
Calculating the derivatives:
- \( \frac{d}{dx}(e^x) = e^x \)
- \( \frac{d}{dx}(x + 1) = 1 \)

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

### Step 5: Evaluate the second derivative at \( x = -1 \)
Now we evaluate the second derivative at the critical point \( x = -1 \):
\[ \frac{d^2y}{dx^2} \bigg|_{x=-1} = e^{-1} (-1 + 2) = e^{-1} (1) = \frac{1}{e} \]

### Step 6: Determine the nature of the critical point
Since \( \frac{d^2y}{dx^2} \bigg|_{x=-1} = \frac{1}{e} > 0 \), this indicates that the function has a local minimum at \( x = -1 \).

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

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