The function you've provided is \( f(x) = e^{x^2} + x \). Here’s some information and insights regarding it:

### Key Features:
1. **Derivative (rate of change)**: 
   To find the derivative of the function, apply the chain rule for the exponential part and the derivative of the linear term:
   \[
   f'(x) = \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(x)
   \]
   The derivative of \( e^{x^2} \) is \( e^{x^2} \cdot 2x \), and the derivative of \( x \) is 1. So:
   \[
   f'(x) = 2xe^{x^2} + 1
   \]

2. **Critical points (extrema)**:  
   These occur where \( f'(x) = 0 \):
   \[
   2xe^{x^2} + 1 = 0
   \]
   Since \( e^{x^2} \) is always positive, solving \( 2xe^{x^2} = -1 \) leads to no real solutions, as this would require \( x \) to be negative and make \( e^{x^2} \) negative, which is impossible. Therefore, there are no critical points.

3. **Concavity (second derivative)**:  
   To explore the concavity, take the second derivative of the function:
   \[
   f''(x) = \frac{d}{dx}(2xe^{x^2}) + \frac{d}{dx}(1) = 2e^{x^2} + 4x^2e^{x^2}
   \]
   This is always positive for all \( x \), indicating that the function is concave up everywhere, meaning it's an increasing function.

4. **Behavior as \( x \to \infty \) or \( x \to -\infty \)**:  
   As \( x \to \infty \), \( e^{x^2} \) grows very rapidly, so \( f(x) \to \infty \).  
   As \( x \to -\infty \), \( e^{x^2} \) also grows rapidly (since \( x^2 \) is positive even for negative \( x \)), and thus \( f(x) \to \infty \) as well.

5. **Asymptotes**:  
   There are no horizontal or vertical asymptotes since the function increases without bound in both directions.

Would you like further details or have any questions?

Here are some related questions for exploration:
1. What is the Taylor series expansion of \( e^{x^2} + x \) around \( x = 0 \)?
2. Can you approximate the value of the function at specific points, such as \( x = 1 \) or \( x = -1 \)?
3. What is the graph of \( f(x) = e^{x^2} + x \) and how does it behave at large \( x \)?
4. How would the derivative \( f'(x) = 2xe^{x^2} + 1 \) be used to find tangent lines at specific points?
5. Can you integrate \( e^{x^2} + x \) with respect to \( x \)?

**Tip:** Exponential functions like \( e^{x^2} \) grow faster than polynomials, so the function’s rate of increase becomes very steep as \( x \) increases.

Explore the mathematical analysis of the function e^(x^2) + x. This solution includes a detailed explanation of its derivative, critical points, concavity, and behavior at infinity. Suitable for university-level calculus students.

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