The problem asks for the value of $x$ at which the function $f(x) = e^{x^4 / 4 + 10px}$ has a critical point. Here's how we can approach it:

Steps to Solve:

1. Find the derivative of $f(x)$. A critical point occurs where the first derivative $f'(x)$ equals zero or does not exist.

   • The function is $f(x) = e^{g(x)}$, where $g(x) = \frac{x^4}{4} + 10px$.
   • Use the chain rule for differentiation: $f'(x) = e^{g(x)} \cdot g'(x)$
   • Now, differentiate $g(x)$: $g'(x) = \frac{d}{dx}\left( \frac{x^4}{4} + 10px \right) = x^3 + 10p$
   • Therefore, the derivative of $f(x)$ is: $f'(x) = e^{\left(\frac{x^4}{4} + 10px\right)} \cdot (x^3 + 10p)$

2. Set the derivative equal to zero to find the critical points: $e^{\left(\frac{x^4}{4} + 10px\right)} \cdot (x^3 + 10p) = 0$ Since $e^{\left(\frac{x^4}{4} + 10px\right)}$ is never zero (because the exponential function is always positive), we can simplify this to: $x^3 + 10p = 0$ Solving for $x$: $x^3 = -10p \quad \Rightarrow \quad x = -\sqrt[3]{10p}$

Conclusion:

The critical point occurs at $x = -\sqrt[3]{10p}$.

Would you like a deeper explanation of any step, or do you have further questions?

Related Questions:

1. What is the second derivative of $f(x)$ at the critical point?
2. How do you classify the critical point as a maximum, minimum, or saddle point?
3. What happens to the critical point when $p = 0$?
4. Can the function have more than one critical point if we modify $f(x)$?
5. How does the parameter $p$ affect the location of the critical point?

Tip:

When solving for critical points in exponential functions, the exponential factor can be disregarded when finding where the derivative is zero because the exponential function never equals zero.