Is the zero function a differentiable, continuous and polynomial function?

Question

For $f(x): \mathbb{R}\rightarrow\mathbb{R}$

1. Differentiable, as the derivative will always be 0
2. Continuous, as it is just a horizontal line with no breaks
3. Polynomial, as it can be written as $f(x) = 0 = (1x^n - 1x^n)$

Would any of my reasoning be wrong?

Answer 1

(1) Correct, $\lim_{x \to a} \frac{f(x) - f(a)}{x-a} = \lim_{x \to a} \frac{0}{x-a} = 0$ for any $a \in \mathbb{R}$

(2) This is not a mathematical reasoning, but rather an intuition why the statement is true. Here's a mathematical proof:

Let $a \in \mathbb{R}, \epsilon > 0$. Choose $\delta =$ "your favorite non-zero positive number". Then, for $x \in \mathbb{R}$ satisfying $|x-a| < \delta$, we have $0 = |f(x)-f(a)| < \epsilon$. Hence $f$ is continuous on its domain.

(3) Correct: Other examples:

$0 = 0x^0, 0 = 0x, \dots$

Answer 2

You are correct in all three statements.

The function $f(x)=0$ is a special case of $f(x)=c$ where $c$ is a constant.

The same statements are true for $f(x)=c$ for any constant $c$ which are considered as polynomials of degree $0$

$1)$ Differentiable, as the derivative will always be $ 0$

$2)$ Continuous, as it is just a horizontal line with no breaks

$3)$ Polynomial, as it can be written as$ f(x) = c+ 0x+0x^2 $