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The problem asks for the value of \( x \) where the function \( f(x) = e^{x^{10} / 10 + 11px} \) has a critical point. To find the critical point, we need to compute the derivative \( f'(x) \), set it equal to zero, and solve for \( x \).

### Steps:
1. **Compute the first derivative \( f'(x) \)**:
   The function is \( f(x) = e^{g(x)} \), where \( g(x) = \frac{x^{10}}{10} + 11px \).
   Using the chain rule:
   \[
   f'(x) = e^{g(x)} \cdot g'(x)
   \]
   Now, find \( g'(x) \):
   \[
   g'(x) = \frac{d}{dx} \left( \frac{x^{10}}{10} + 11px \right) = x^9 + 11p
   \]

   Therefore, the derivative becomes:
   \[
   f'(x) = e^{\frac{x^{10}}{10} + 11px} \cdot (x^9 + 11p)
   \]

2. **Set \( f'(x) = 0 \)**:
   Since \( e^{\frac{x^{10}}{10} + 11px} \neq 0 \) for any real \( x \), the critical points occur when:
   \[
   x^9 + 11p = 0
   \]

3. **Solve for \( x \)**:
   \[
   x^9 = -11p
   \]
   \[
   x = (-11p)^{1/9}
   \]

Thus, the function has a critical point when \( x = (-11p)^{1/9} \).

Would you like more details or have any further questions? Here are 5 related questions to consider:

1. What is the definition of a critical point in calculus?
2. How does the chain rule apply when differentiating exponential functions?
3. Can \( p \) be any value, and how does it affect the critical point?
4. What does the behavior of the second derivative tell us about the nature of this critical point?
5. How would this problem change if \( f(x) \) involved different exponents or terms?

**Tip**: To analyze critical points thoroughly, always check both the first and second derivatives for sign changes and concavity.

For what value of x does the function f(x) = e^(x^10 / 10 + 11px) have a critical point?

Learn how to find the critical point of the function f(x) = e^(x^10 / 10 + 11px) by computing the first derivative and solving for x. This solution covers differentiation using the chain rule and solving for critical points in calculus.

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