The statement is True. Here's why:

If $f(x)$ is an $n$-degree polynomial, it means that the highest power of $x$ in the polynomial is $x^n$. The $n$-degree polynomial has the general form:

$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$

Each differentiation of $f(x)$ reduces the degree of the polynomial by 1:

• The first derivative $f'(x)$ is a polynomial of degree $n-1$.
• The second derivative $f''(x)$ is a polynomial of degree $n-2$.
• ...
• The $n$-th derivative $f^{(n)}(x)$ is a constant (since the derivative of the highest degree term, $a_n x^n$, becomes $a_n \cdot n$, and each subsequent term reduces to zero).

If you differentiate this constant once more (i.e., take the $(n+1)$-th derivative), the result will be zero:

$f^{(n+1)}(x) = 0$

Thus, the $(n+1)$-th derivative of an $n$-degree polynomial is always zero.

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Would you like further details or have any other questions?

Here are 5 related questions to deepen your understanding:

1. What is the second derivative of a quadratic polynomial?
2. Can the degree of a polynomial change after differentiation?
3. How does the derivative of a linear function behave?
4. What is the geometric interpretation of the second derivative?
5. How do higher-order derivatives relate to the curvature of a graph?

Tip: The degree of a polynomial decreases with each differentiation, until it eventually becomes zero after enough differentiations!