Can there be a function, that with the slope $f'(x_0) = 0$ at the single root $x_0$

Question

Let $f$ be a function of the form $$ f(x) = \sum_{i = 0}^n a_i x^i = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x+ a_0 $$ with $a_n, a_{n - 1}, ..., a_1,a_0 \in \mathbb{R}$.

Let $x_o$ be a single root of $f$ $$f(x_0) = 0 \iff f(x) = (x - x_0) * p(x)$$

Is ist true that there is no (polynomial) function with the slope of zero at $P(x_0/f(x_o))$? $$ f'(x_0) = 0 $$

My approaches so far:

Proof for any function of the form $f(x) = ax + b$, with $a \in \mathbb{R} \setminus 0, b \in \mathbb{R}$ $$ f'(x) = a \iff a \neq 0 $$

Proof for any function of the form $f(x) = ax^2 + bx + c$, with $a \in \mathbb{R} \setminus 0,b,c \in \mathbb{R}$ $$ x_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ f'(x) = 2ax + b $$ $$ f'(x_{1/2}) = 2a\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} + b = \pm \sqrt{b^2 - 4ac} $$ if $\sqrt{b^2 - 4ac} = 0$, then ther is no a double, not a single zero

Proof for any function of the form $f(x) = (x - x_1)(x - x_2)(x - x_3)$

$$ f'(x) = ((x -x_1) + (x - x_2)) * (x - x_3) + (x - x_1)(x - x_2) $$

$f'(x_1) = (x -x_2) * (x - x_3)$, $f'(x_2) = (x-x_1) * (x - x_3)$, $f'(x_3) =(x - x_1)(x - x_2) $

All three are zero under the premise $x_1 \neq x_2 \neq x_3$ (single root)

I am looking for a way of showing this for any polynomial function!

score 3 · Answer 1 · answered Jan 08 '16 at 22:11

3

Since $f$ factors as $f(x) = (x - x_0) p(x)$ with $p(x_0) \ne 0$, notice that

$$f'(x) = (x - x_0) p'(x) + p(x)$$

so that the derivative of $f'$ at $x_0$ is $p(x_0)$.

answered Jan 08 '16 at 22:11

And how can you show, that $p(x_0) \neq 0$ (It must be of course, otherwise $x_0$ wouldn't be a single zero) – Entimon Jan 08 '16 at 22:17
@Entimon If $p(x_0) = 0$, then $p$ factors as $(x - x_0) q(x)$, which leads to $f(x) = (x - x_0)^2 q(x)$. As a generalization, you can show that you have a root of multiplicity $n$ when the first $n$ derivatives are zero. – Jan 08 '16 at 22:18
For what multiplicities of a root does this apply to (only single roots, or odd roots ...)? – Entimon Jan 08 '16 at 22:28
@Entimon Any of them. – Jan 08 '16 at 22:29
What about $f(x) = x^2$ with $x_0 = 0$ and $f'(x_0) = 0$? – Entimon Jan 08 '16 at 22:32
@Entimon: What about it? That's a double zero. – Brian Tung Jan 08 '16 at 22:34
Yes, that factors as $(x - 0)^2$.... – Jan 08 '16 at 22:34
Yeah, so $f'(x_0) \neq 0$, if it's an single zero? Or does this apply also to e.g. a triple zero? – Entimon Jan 08 '16 at 22:46
"Not a single zero" includes triple zeros. – Jan 08 '16 at 22:46
So there is no $f(x) = (x - x_0)^3 * p(x)$ with $f'(x) \neq 0$??? – Entimon Jan 08 '16 at 22:48
No, there is not (and the answer shows why). – Jan 08 '16 at 22:49

Can there be a function, that with the slope $f'(x_0) = 0$ at the single root $x_0$

1 Answers1