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The given problem states the following function family:

\[
f_k(x) = kx^4 - x^2, \quad k \in \mathbb{R}
\]

You are asked to:

1. **Find all the zeros** of \( f_k(x) \).
2. **Determine the extrema** of \( f_k(x) \) and their nature.
3. **Find the inflection points** of \( f_k(x) \), all in terms of \( k \).

### 1. **Zeros of \( f_k(x) \)**

We solve for \( f_k(x) = 0 \):

\[
k x^4 - x^2 = 0
\]

Factor out \( x^2 \):

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

Thus, \( x^2 = 0 \) or \( kx^2 - 1 = 0 \). This gives us two cases:

- \( x = 0 \) is one root.
- \( kx^2 - 1 = 0 \implies x^2 = \frac{1}{k} \). This results in:

\[
x = \pm \frac{1}{\sqrt{k}}
\]

So the zeros depend on \( k \). If \( k > 0 \), we get three distinct real roots: \( x = 0, \pm \frac{1}{\sqrt{k}} \). If \( k \leq 0 \), the only real zero is \( x = 0 \), since for \( k < 0 \), \( \frac{1}{k} \) is negative and does not yield real roots.

### 2. **Extrema of \( f_k(x) \)**

To find the extrema, we compute the first derivative of \( f_k(x) \):

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

Set the derivative equal to zero to find critical points:

\[
4k x^3 - 2x = 0 \implies 2x(2k x^2 - 1) = 0
\]

This gives two cases:

- \( x = 0 \)
- \( 2k x^2 - 1 = 0 \implies x^2 = \frac{1}{2k} \implies x = \pm \frac{1}{\sqrt{2k}} \) (valid only if \( k > 0 \)).

Thus, the critical points are \( x = 0 \) and \( x = \pm \frac{1}{\sqrt{2k}} \) for \( k > 0 \).

To classify these points, we use the second derivative:

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

- At \( x = 0 \): \( f_k''(0) = -2 \), so \( x = 0 \) is a local maximum.
- At \( x = \pm \frac{1}{\sqrt{2k}} \): \( f_k''\left( \pm \frac{1}{\sqrt{2k}} \right) = 12k \cdot \frac{1}{2k} - 2 = 4 - 2 = 2 > 0 \), so these points are local minima.

### 3. **Inflection Points**

To find inflection points, we set the second derivative to zero:

\[
f_k''(x) = 12k x^2 - 2 = 0 \implies x^2 = \frac{1}{6k} \implies x = \pm \frac{1}{\sqrt{6k}}
\]

These are the inflection points, valid only when \( k > 0 \).

---

### Summary:

- **Zeros**: \( x = 0 \) and \( x = \pm \frac{1}{\sqrt{k}} \) (for \( k > 0 \)).
- **Extrema**:
  - \( x = 0 \): local maximum.
  - \( x = \pm \frac{1}{\sqrt{2k}} \): local minima (for \( k > 0 \)).
- **Inflection points**: \( x = \pm \frac{1}{\sqrt{6k}} \) (for \( k > 0 \)).

Would you like more details on any part of the solution?

---

### Related Questions:

1. How does the behavior of the function change when \( k = 0 \)?
2. What happens to the number of extrema as \( k \to 0 \)?
3. How do the inflection points behave as \( k \) increases?
4. Can you solve the problem for specific values of \( k \), like \( k = 1 \) or \( k = -1 \)?
5. How would the graph of \( f_k(x) \) look for different values of \( k \)?

### Tip: 
When analyzing functions depending on a parameter, it helps to sketch the graph for different values of the parameter to visualize changes in behavior.

Gegeben ist die Funktionsschar 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.

Learn how to find the zeros, extrema, and inflection points for the function family f_k(x) = kx^4 - x^2. This step-by-step guide covers derivatives and critical points analysis, suitable for university-level calculus students.

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