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We are given the function:

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

and asked to find for what value of \( x \) the function \( f(x) \) has a critical point, and express the result in terms of \( p \).

### Steps to Solve:
1. **Critical points** occur where the first derivative of the function \( f(x) \) equals zero.
   
   We need to compute \( f'(x) \).

2. **Derivative of the given function**:
   - The first term, \( e^{x^4/4} \), will be differentiated using the chain rule.
   - The second term, \( 10px \), is linear, and its derivative is straightforward.

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

For the first term:
\[
\frac{d}{dx} \left( e^{x^4/4} \right) = e^{x^4/4} \cdot \frac{d}{dx} \left( \frac{x^4}{4} \right) = e^{x^4/4} \cdot x^3
\]

For the second term:
\[
\frac{d}{dx} \left( 10px \right) = 10p
\]

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

3. **Set the derivative equal to zero** to find critical points:
\[
e^{x^4/4} \cdot x^3 + 10p = 0
\]

4. **Solve for \( x \)**:
   \[
   e^{x^4/4} \cdot x^3 = -10p
   \]
   Since \( e^{x^4/4} \) is always positive, it follows that the sign of \( x^3 \) must match the sign of \( -10p \). Dividing both sides by \( e^{x^4/4} \), we get:
   \[
   x^3 = \frac{-10p}{e^{x^4/4}}
   \]

Thus, the critical points are determined by the above equation. We cannot explicitly solve this for \( x \) without further information about \( p \), but this equation provides the implicit relationship between \( x \) and \( p \).

---

Would you like more details on how to interpret this or help with other steps? 

### Related Questions:
1. How do we determine if this critical point is a maximum, minimum, or saddle point?
2. Can we further approximate \( x \) for specific values of \( p \)?
3. What happens to the critical points when \( p = 0 \)?
4. How does the nature of \( e^{x^4/4} \) affect the solution for \( x \)?
5. What is the second derivative of \( f(x) \), and how does it help in identifying the concavity?

**Tip:** The nature of critical points depends not only on the first derivative but also on the second derivative (the concavity test).

For what value of x does the function f(x) = e^(x^4/4) + 10px have a critical point? Answer with an expression that includes p.

Learn how to find the critical points of the function f(x) = e^(x^4/4) + 10px by setting the derivative equal to zero and solving for x in terms of p. This solution involves applying the chain rule and solving for critical points.

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