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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?

glockm15
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2 Answers2

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(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:

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

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    Nitpick: For many authors, $0x^0$ is not defined at $x=0.$ – Dave L. Renfro Apr 11 '18 at 12:04
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    @DaveL.Renfro: For many authors, it is, contrary to your claim. See the standard form of the Taylor series, the binomial theorem, Elementary Function Arithmetic, ... Defining $x^0 = 1$ for real $s$ is perfectly in line with the elegant viewpoint that natural exponents correspond to iterates of multiplication. – user21820 Apr 21 '18 at 12:10
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    @DaveL.Renfro In the context of polynomials, which is what (3) is concerned with, $x^0$ is definitely $1$, hence $0x^0=0$ for every $x$ in the underlying field $K$. How would one write every polynomial $P(x)$ in $K[x]$ as $$P(x)=\sum_{i=0}^na_ix^i$$ otherwise? And the associated polynomial function $f_P:K\to K$ is indeed defined by $$f_P(t)=\sum_{i=0}^na_it^i=a_0+a_1t+a_2t^2+\cdots+a_nt^n$$ for every $t$ in $K$, it seems to me. – Did May 02 '18 at 17:01
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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 $

Mohammad Riazi-Kermani
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