If $p(x) = x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of $f$ around $x=1$, then $1$ is a local minimum

Question

Consider $f:\mathbb{R} \to \mathbb{R} \in C^4$. Show that $p(x) = x^4-4x^3+6x^2-4x+1$ is the Taylor polynomial of order $4$ of $f$ around $x=1$, then $1$ is a local minimum.

I'm not sure how to proceed. I know that $p(x) = \sum\limits_{k=0}^{4} \dfrac{f^{(n)}(1)(x-1)^n}{n!} = f(1)+f'(1)(x-1)+\dfrac{f''(1)(x-1)^2}{2} + \dfrac{f'''(1)(x-1)^{3}}{6} + \dfrac{f^{(4)}(1)(x-1)^4}{24}=x^4-4x^3+6x^2-4x+1$

and I can see that $p(1) = 0 \implies f(1)=0$. But how can I obtain information about $f'(1)$ and $f''(1)$?

Answer 1

For $0 \le k \le 4$, let $f^{(k)}$ denote the $k$-th derivative of $f$.

Since the Taylor polynomial of degree $4$ for $f$ at $x=1$ is $$x^4-4x^3+6x^2-4x+1=(x-1)^4$$ it follows that $f^{(k)}(1)=0$, for $0\le k\le 3$, and $f^{(4)}(1)=24$.

For brevity, let

• "holds near $x = 1$ on the left" mean "holds in some open interval with right endpoint at $1$".$\\[4pt]$
• "holds near $x = 1$ on the right" mean "holds in some open interval with left endpoint at $1$".$\\[4pt]$
• "holds near $x = 1$" mean "holds near $x=1$ on the left" and "holds near $x=1$ on the right".

Since $f^{(4)}$ is continuous and $f^{(4)}(1) > 0$, it follows that $f^{(4)}(x) > 0$ near $x=1$.

Using the above result and applying the Mean Value Theorem, it follows that $f^{(3)}(x) < f^{(3)}(1) = 0$ near $x=1$ on the left, and $f^{(3)}(x) > f^{(3)}(1) = 0$ near $x=1$ on the right.

Using the above result and applying the Mean Value Theorem, it follows that $f^{(2)}(x) > f^{(2)}(1)=0$ near $x=1$.

Using the above result and applying the Mean Value Theorem, it follows that $f^{(1)}(x) < f^{(1)}(1) = 0$ near $x=1$ on the left, and $f^{(1)}(x) > f^{(1)}(1) = 0$ near $x=1$ on the right.

Using the above result and applying the Mean Value Theorem, it follows that $f(x)=f^{(0)}(x) > f^{(0)}(1)=0$ near $x=1$.

Therefore $f$ has a local minimum at $x=1$.

To show how the Mean Value Theorem was applied, let's examine one of the cases . . .

Since $f^{(4)}(x) > 0$ near $x=1$, we have $f^{(4)}(x) > 0$ for all $x\in (1,b)$, for some $b > 1$.

Suppose $f^{(3)}(t) \le 0$, for some $t\in (1,b)$.

Then by the MVT, it would follow that $$f^{(4)}(s)=\frac{f^{(3)}(t)-f^{(3)}(1)}{t-1}\le 0$$

for some $s\in (1,t)$, contradiction, since $s\in (1,t)$ implies $s\in (1,b)$.

Therefore $f^{(3)}(x) > 0$ for all $x\in (1,b)$.

The reasoning for the other cases is analogous.

Answer 2

By the Remainder theorem of the Taylor polynomials we have $$ f(x)=P(x) + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$

$$ = (x-1)^4 + \frac {f^{(5)}(\xi)}{5!} (x-1)^5$$

$$ = (x-1)^4 \big[1+\frac {f^{(5)}(\xi)}{5!} (x-1)\big]$$

Let $x$ be close enough to $1$ such that $$ \big| \frac {f^{(5)}(\xi)}{5!} (x-1)\big|<1/2$$ Then $$ \frac {(x-1)^4}{2}<f(x)<\frac{3(x-1)^4}{2}$$

Thus $f(x)$ has a minimum at $x=1$.