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To find the critical points of a function, we need to calculate the derivative of the function \( f(x) \) and set it equal to zero. Let's break this problem down:

Given:
\[
f(x) = e^{\frac{x^4}{4} + 10px}
\]

### Step 1: Find the derivative \( f'(x) \)
The function \( f(x) \) is of the form \( e^{g(x)} \), where \( g(x) = \frac{x^4}{4} + 10px \). By the chain rule, the derivative of \( f(x) \) is:

\[
f'(x) = e^{g(x)} \cdot g'(x)
\]

Now, we need to compute \( g'(x) \), the derivative of \( g(x) \):
\[
g'(x) = \frac{d}{dx} \left( \frac{x^4}{4} + 10px \right) = x^3 + 10p
\]

Thus, the derivative of \( f(x) \) becomes:
\[
f'(x) = e^{\frac{x^4}{4} + 10px} \cdot (x^3 + 10p)
\]

### Step 2: Find the critical points
To find the critical points, we set \( f'(x) = 0 \). Since the exponential term \( e^{\frac{x^4}{4} + 10px} \) is never zero, we only need to solve:

\[
x^3 + 10p = 0
\]

This simplifies to:
\[
x^3 = -10p
\]

Taking the cube root of both sides gives:
\[
x = \sqrt[3]{-10p}
\]

### Final Answer:
The value of \( x \) at which the function has a critical point is \( x = \sqrt[3]{-10p} \).

Let me know if you'd like more details or if you have any further questions!

Here are 5 related questions:
1. How do critical points help in determining the behavior of a function?
2. What is the importance of using the chain rule in finding derivatives?
3. How can you determine if the critical point is a maximum or minimum?
4. What happens when \( p = 0 \) in this problem?
5. How do higher-degree polynomials affect the number of critical points?

**Tip:** Always double-check if your critical points lead to local maxima or minima by evaluating the second derivative!

For what value of x does the function f(x) have a critical point if f(x) = e^(x^4/4 + 10px)?

Learn how to find the critical points of the function f(x) = e^(x^4/4 + 10px) by calculating its derivative using the chain rule. Understand how to solve for the value of x when the derivative equals zero. Suitable for undergraduate calculus students.

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