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The given problem states:

"Gegeben ist die Funktionenschar \( f_k: x \mapsto kx^4 - x^2 \), \( k \in \mathbb{R} \). Bestimmen Sie alle Nullstellen von \( f_k \), Extrema von \( f_k \) und ihre Art sowie Wendestellen von \( f_k \) in Abhängigkeit von \( k \)."

This translates to:

"Given is the family of functions \( f_k: x \mapsto kx^4 - x^2 \), \( k \in \mathbb{R} \). Determine all the zeros of \( f_k \), the extrema of \( f_k \) and their type, as well as the inflection points of \( f_k \), depending on \( k \)."

Let's break this problem down step by step:

### 1. Zeros (Nullstellen) of \( f_k(x) \):
We are tasked to find where \( f_k(x) = 0 \), i.e., solving:

\[
f_k(x) = kx^4 - x^2 = 0
\]

Factoring the equation:

\[
x^2(kx^2 - 1) = 0
\]

Thus, the solutions are:

\[
x = 0 \quad \text{or} \quad kx^2 = 1 \Rightarrow x^2 = \frac{1}{k} \quad \text{(for } k \neq 0\text{)}.
\]

This gives:

\[
x = \pm \frac{1}{\sqrt{k}} \quad \text{for } k > 0.
\]

For \( k < 0 \), there are no real solutions from \( x^2 = \frac{1}{k} \).

So the zeros are:
- \( x = 0 \) (for all \( k \)),
- \( x = \pm \frac{1}{\sqrt{k}} \) (only for \( k > 0 \)).

### 2. Extrema of \( f_k(x) \):
To find the extrema, we compute the first derivative \( f_k'(x) \):

\[
f_k'(x) = 4kx^3 - 2x
\]

Setting \( f_k'(x) = 0 \) for critical points:

\[
4kx^3 - 2x = 0 \quad \Rightarrow \quad x(4kx^2 - 2) = 0
\]

Thus:

\[
x = 0 \quad \text{or} \quad 4kx^2 = 2 \Rightarrow x^2 = \frac{1}{2k}.
\]

So the critical points are:

\[
x = 0 \quad \text{or} \quad x = \pm \frac{1}{\sqrt{2k}} \quad (k > 0).
\]

### 3. Inflection Points (Wendestellen):
The inflection points occur where the second derivative \( f_k''(x) \) changes sign. First, we compute \( f_k''(x) \):

\[
f_k''(x) = 12kx^2 - 2.
\]

Setting \( f_k''(x) = 0 \) for potential inflection points:

\[
12kx^2 = 2 \quad \Rightarrow \quad x^2 = \frac{1}{6k}.
\]

So the possible inflection points are:

\[
x = \pm \frac{1}{\sqrt{6k}} \quad (k > 0).
\]

For \( k < 0 \), there are no real inflection points.

### Conclusion:
- **Zeros**: \( x = 0 \), and \( x = \pm \frac{1}{\sqrt{k}} \) for \( k > 0 \).
- **Extrema**: \( x = 0 \) (global max/min depending on \( k \)), and \( x = \pm \frac{1}{\sqrt{2k}} \) for \( k > 0 \).
- **Inflection Points**: \( x = \pm \frac{1}{\sqrt{6k}} \) for \( k > 0 \).

Would you like further details or any clarifications?

Here are 5 related questions:
1. How does the behavior of the function change for negative \( k \)?
2. What is the nature of the extrema at \( x = 0 \)?
3. How do the inflection points impact the curvature of the graph?
4. What happens to the zeros as \( k \) approaches 0?
5. Can you graph the function for different values of \( k \)?

**Tip**: Always check the second derivative to confirm the nature (max/min) of critical points!

Gegeben ist die Funktionenschar f_k: x ↦ kx^4 - x^2, k ∈ R. Bestimmen Sie alle Nullstellen von f_k, Extrema von f_k und ihre Art sowie Wendestellen von f_k in Abhängigkeit von k.

Solve the polynomial function f_k(x) = kx^4 - x^2 by determining its zeros, extrema, and inflection points. This problem involves derivatives and critical point analysis, making it suitable for high school calculus students.

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