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The task asks to prove that the function \( F(x) = x^6 - 3x + 2 \) is a **primitive (antiderivative)** of the function \( f(x) = 6x^5 - 3 \) on the interval \( (-\infty, +\infty) \).

### Steps to Prove:

To prove that \( F(x) \) is the primitive (antiderivative) of \( f(x) \), we need to take the derivative of \( F(x) \) and verify that it equals \( f(x) \).

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

2. Compute the derivative of \( F(x) \) with respect to \( x \):
   \[
   F'(x) = \frac{d}{dx}(x^6) - \frac{d}{dx}(3x) + \frac{d}{dx}(2)
   \]
   - The derivative of \( x^6 \) is \( 6x^5 \).
   - The derivative of \( -3x \) is \( -3 \).
   - The derivative of the constant \( 2 \) is \( 0 \).

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

3. Since the derivative of \( F(x) \) equals \( f(x) = 6x^5 - 3 \), we have proven that \( F(x) \) is the antiderivative of \( f(x) \) on \( (-\infty, +\infty) \).

Would you like more details or further clarifications?

### Related Questions:
1. How do you compute the antiderivative of a more complex polynomial function?
2. What is the role of the constant when computing an indefinite integral?
3. Can a function have more than one primitive (antiderivative)?
4. How do you verify if a function is a primitive for a given interval?
5. What is the geometric interpretation of an antiderivative?

### Tip:
When taking the derivative of a polynomial, remember that the derivative of a constant is always zero.

Доведіть, що функція F = x^6 - 3x + 2 є первісною для функції f = 6x^5 - 3 на (-∞; +∞).

Prove that the function F(x) = x^6 - 3x + 2 is an antiderivative of f(x) = 6x^5 - 3 over the interval (-∞, +∞). This problem involves applying derivative rules and understanding the concept of primitives in calculus. Suitable for students in Grades 11-12.

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