I have a random function $f(x)$. I know nothing about the shape of the function, but I know

  • $f'(x)$ exists, $x \in \mathbb{R}$
  • $f(3)=1$, $f'(3)=5$, $f''(3)=7, f''''(3)=3$
  • If $x\in (3,6)$, then $f'(x) \in[1,5]$
  • $-3\leq f''(x)\leq 9, \forall x \in \mathbb{R}$
  • $-4 \leq f'''(x) \leq -1, \forall x \in \mathbb{R}$

So having this information, how do I determine the second degree taylor polynomial, $T_{2, 3}$ for this function? I know the value of $f'(3)$ up to $f''''(3)$, which is perfect for this question. But the taylor polynomial theorem requires the function is differentiable everywhere in order for this to work right? Do I need to prove that first, and how should I prove it?

Also, how do we find the upper bound for the error using $T_{2,3}$ for f(5)? Using taylor's remainder theorem is fine but which points should I choose for $f'''(c)$?

  • $\begingroup$ comment: It is more common to use notation $f^{(4)}(x)$ instead of $f''''(x)$ to denote $4th$ derivative of $f(x)$ $\endgroup$
    – Etemon
    Dec 9 '20 at 23:40
  • $\begingroup$ That is true. Also ignore "If x∈(3,6), then f′(x)∈[1,5]", this does not have much to do with the question, as I figured out. $\endgroup$ Dec 10 '20 at 0:11
  • $\begingroup$ Can someone plz help me with this question? $\endgroup$ Dec 10 '20 at 4:38

Second degree taylor polynomial: $1+5(x-3)+\frac{7}{2}(x-3)^2$

Upper bound on error for $f(5)$: $\frac{16}{3}$

We already know the function is differentiable everywhere, because $f'(x)$ exists $x \in R$

Because the function is differentiable, we can write out its Taylor series expansion (centered at 3 because that's where we know the derivatives, and up to a degree of 2 because we're only trying to find the second degree Taylor polynomial).


To find the upper bound on the error, we use Taylor's Remainder Formula:

if $\left|f^{(k+1)}(z)\right|\leq M$ for all z between x and a, then $$\left|R_k(x)\right|\leq \frac{M}{(k+1)!}|x-a|^{k+1}$$ Where $R_x$ is the error in the approximation

So in our case $k=2$, $\left|f^{(3)}(z)\right|\leq M = 4$ (because $-4 \leq f'''(x) \leq -1, \forall x \in \mathbb{R}$), $x=5$, and $a=3$, so $$\left|R_2(x)\right|\leq \frac{4}{(2+1)!}|5-3|^{2+1}=\frac{16}{3}$$


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