Determine all the critical points for the function : `f(x)=xe^(x^2)`

Answer

To determine all the critical points for the function \( f(x) = x e^{x^2} \), we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first need to compute the derivative \( f'(x) \).

Given:
\[ f(x) = x e^{x^2} \]

Using the product rule, where \( u = x \) and \( v = e^{x^2} \):
\[ f'(x) = u'v + uv' \]
Here, \( u' = 1 \) and \( v' = e^{x^2} \cdot 2x \) (using the chain rule).

Thus,
\[ f'(x) = 1 \cdot e^{x^2} + x \cdot (e^{x^2} \cdot 2x) \]
\[ f'(x) = e^{x^2} + 2x^2 e^{x^2} \]
Factoring out \( e^{x^2} \):
\[ f'(x) = e^{x^2} (1 + 2x^2) \]

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

### Step 3: Analyze the factors
The term \( e^{x^2} \) is always greater than 0 for all real \( x \) (since the exponential function never equals zero). Therefore, we only need to consider the second factor:
\[ 1 + 2x^2 = 0 \]

### Step 4: Solve for \( x \)
Solving \( 1 + 2x^2 = 0 \):
\[ 2x^2 = -1 \]
This equation has no real solutions since \( x^2 \) cannot be negative.

### Conclusion
Since there are no values of \( x \) that satisfy \( f'(x) = 0 \), we conclude that there are no critical points for the function \( f(x) = x e^{x^2} \).