https://file.aiquickdraw.com/mathbot/file/17290600481680pmkgxpr.jpg

To find the critical points of the function \( f(x) = e^{\frac{x^4}{4} + 10px} \), we need to determine where the derivative \( f'(x) \) is equal to zero.

### Step 1: Differentiate \( f(x) \)

We apply the product rule and chain rule to differentiate \( f(x) \). Let:
\[
u = \frac{x^4}{4} + 10px
\]
Thus:
\[
f(x) = e^u
\]
The derivative of \( f(x) \) is:
\[
f'(x) = e^u \cdot u'
\]
where \( u' \) is the derivative of \( u \).

### Step 2: Differentiate \( u \)

\[
u = \frac{x^4}{4} + 10px
\]
\[
u' = x^3 + 10p
\]

### Step 3: Set \( f'(x) = 0 \)

\[
f'(x) = e^{\frac{x^4}{4} + 10px} \cdot (x^3 + 10p)
\]

For \( f'(x) = 0 \), we need \( x^3 + 10p = 0 \) (since \( e^u \) is never zero).

\[
x^3 + 10p = 0
\]
\[
x^3 = -10p
\]
\[
x = \sqrt[3]{-10p}
\]

### Conclusion

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

Let me know if you'd like any further explanation or have other questions!

### Follow-up Questions
1. What are the types of critical points (local minima, maxima, or saddle points) for this function?
2. How can the second derivative be used to classify the critical point found?
3. What happens to the critical point when \( p = 0 \)?
4. Can you find the inflection points of this function?
5. How would the function's graph look around the critical point?

### Tip
To determine if a critical point is a minimum, maximum, or neither, the second derivative test is often useful.

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

Solve for the critical point of the function f(x) = e^(x^4/4 + 10px) by differentiating and finding where the derivative equals zero. Explore calculus concepts such as differentiation and critical points. Suitable for undergraduate calculus students.

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