The problem asks to find the value of $x$ for which the function $f(x) = e^{x^4/4 + 10px}$ has a critical point, and the answer should include $p$.

Step 1: Find the first derivative of $f(x)$

To locate critical points, we need to take the derivative of $f(x)$ and set it equal to zero. First, differentiate the function $f(x)$.

$f(x) = e^{x^4/4 + 10px}$

Using the chain rule, the derivative is:

$f'(x) = e^{x^4/4 + 10px} \cdot \frac{d}{dx}\left(\frac{x^4}{4} + 10px\right)$

Now differentiate the exponent $\frac{x^4}{4} + 10px$:

$\frac{d}{dx}\left(\frac{x^4}{4} + 10px\right) = x^3 + 10p$

Thus, the first derivative becomes:

$f'(x) = e^{x^4/4 + 10px} \cdot (x^3 + 10p)$

Step 2: Set the derivative equal to zero

To find the critical points, set $f'(x) = 0$:

$e^{x^4/4 + 10px} \cdot (x^3 + 10p) = 0$

Since the exponential function is never zero, we have:

$x^3 + 10p = 0$

Step 3: Solve for $x$

Solve the equation $x^3 + 10p = 0$ for $x$:

$x^3 = -10p$

$x = \sqrt[3]{-10p}$

Final Answer:

The value of $x$ for which $f(x)$ has a critical point is:

$x = \sqrt[3]{-10p}$

Would you like more details or have any questions?

Here are 5 follow-up questions related to this problem:

1. What is the significance of a critical point in the context of a function?
2. How would you check if the critical point found is a maximum, minimum, or saddle point?
3. How does the value of $p$ affect the location of the critical point?
4. What happens to the critical point when $p = 0$?
5. How would the second derivative of $f(x)$ help in analyzing the nature of the critical point?

Tip: To find whether the critical point is a maximum or minimum, use the second derivative test by checking the sign of $f''(x)$ at the critical point.