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

Step 1: Differentiate $f(x)$ with respect to $x$

Using the chain rule, the derivative $f'(x)$ is: $f'(x) = \frac{d}{dx}\left( e^{\frac{x^4}{4}} \right) + \frac{d}{dx}(10px)$

1. For the first term, $\frac{d}{dx}\left( e^{\frac{x^4}{4}} \right) = e^{\frac{x^4}{4}} \cdot \frac{d}{dx} \left( \frac{x^4}{4} \right) = e^{\frac{x^4}{4}} \cdot x^3$.

2. The derivative of $10px$ is simply $10p$, since $p$ is treated as a constant.

So, $f'(x) = e^{\frac{x^4}{4}} \cdot x^3 + 10p$

Step 2: Set $f'(x) = 0$

To find the critical points, set $f'(x) = 0$: $e^{\frac{x^4}{4}} \cdot x^3 + 10p = 0$

Since $e^{\frac{x^4}{4}}$ is always positive for real values of $x$, we can divide by it without changing the equation: $x^3 + 10p = 0$

Solving for $x$: $x^3 = -10p$ $x = -\sqrt[3]{10p}$

Answer

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

Would you like further details on this process or have additional questions?

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Here are five related questions to expand your understanding:

1. How would the critical point change if $p$ were negative?
2. What happens to the critical point if $p = 0$?
3. Are there any conditions under which this function $f(x)$ would have no critical points?
4. How can you determine if the critical point is a minimum or maximum?
5. What is the effect of changing the exponent in the term $x^4/4$?

Tip: Remember that a critical point occurs where the derivative equals zero or is undefined—key information for analyzing function behavior.