https://file.aiquickdraw.com/mathbot/file/1727431435324bq9v97ks.jpg

The problem in the image asks to prove that the function \( F(x) = x^6 - 3x + 2 \) is a **primitive** (an antiderivative) for the function \( f(x) = 6x^5 - 3 \) over the interval \( (-\infty, +\infty) \).

### Solution:

To prove that \( F(x) \) is a primitive for \( f(x) \), we need to differentiate \( F(x) \) and show that the derivative is equal to \( f(x) \).

#### Step 1: Differentiate \( F(x) \)
The function \( F(x) = x^6 - 3x + 2 \).

We differentiate term by term:
\[
\frac{d}{dx} (x^6) = 6x^5
\]
\[
\frac{d}{dx} (-3x) = -3
\]
\[
\frac{d}{dx} (2) = 0
\]

Thus, the derivative of \( F(x) \) is:
\[
F'(x) = 6x^5 - 3
\]

#### Step 2: Compare \( F'(x) \) with \( f(x) \)
The function \( f(x) = 6x^5 - 3 \), and from Step 1, we found that \( F'(x) = 6x^5 - 3 \).

Since \( F'(x) = f(x) \), we have proven that \( F(x) \) is indeed a primitive (antiderivative) for \( f(x) \) over \( (-\infty, +\infty) \).

### Conclusion:
The function \( F(x) = x^6 - 3x + 2 \) is a primitive for \( f(x) = 6x^5 - 3 \).

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Do you have any questions or need further details on the solution?

Here are 5 related questions you could explore:
1. What are the conditions for a function to have a primitive?
2. How do you find the general solution to a differential equation using primitives?
3. What is the role of constants when finding antiderivatives?
4. Can a function have more than one antiderivative?
5. How does the Fundamental Theorem of Calculus relate differentiation and integration?

**Tip:** When verifying if a function is a primitive of another, always take the derivative of the potential primitive and compare it with the given function.

Prove that the function F(x) = x^6 - 3x + 2 is a primitive of the function f(x) = 6x^5 - 3 over the interval (-∞, +∞).

Learn how to prove that F(x) = x^6 - 3x + 2 is a primitive of the function f(x) = 6x^5 - 3 over the interval (-∞, +∞) using differentiation. This problem involves key calculus concepts such as derivatives and polynomials, suitable for advanced high school students.

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