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!