How to calculate $f^{\prime\prime}(3)$ from a value table of $f(x)$?

Question

Is there enough information to calculate $f^{\prime\prime}(3)$ from this table?

My intuition says there is a way.

\begin{align} f(3) &= 6\\ f^\prime(3) &= 1\\ &= \lim_{x\to3}\dfrac{f(x)-f(3)}{x-3}\\ f^{\prime\prime} &= \lim_{x\to3}\dfrac{f^\prime(x)-f^\prime(3)}{x-3}\\ &= \lim_{x\to3}\dfrac{f^\prime(x)-1}{x-3}\\ &= \lim_{x\to3}\dfrac{\Bigg(\lim_{x\to3}\dfrac{f(x)-f(3)}{x-3}\Bigg)-1}{x-3}\\ &= \lim_{x\to3}\dfrac{\Bigg(\lim_{x\to3}\dfrac{f(x)-6}{x-3}\Bigg)-1}{x-3}\\ &= \Bigg(\lim_{x\to3}\dfrac{f(x)-6}{(x-3)^2}\Bigg)-\lim_{x\to3}\dfrac{1}{x-3}\\ \end{align}\

and then I get stuck...

Answer 1

With the given information, it is not possible to determine $f''(3)$ exactly. Since only the values of $f$ near $a=3$ are relevant, I will ignore the other information and only consider $f(3)=6$ and $f'(3)=1$. Here are a few examples:

1. Consider $f(x)=x+3$. Then $f(3)=6$, $f'(3)=1$ and $f''(3)=0$.

2. For the quadratic function $f(x)=a(x-3)^2+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(3)=2a$ can be anything.

3. For $f(x)=(x-3)|x-3|+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(x)$ does not exist (see below).

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What can be done in your case is a broad estimation of $f''(3)$ assuming $f$ is "nice enough": the average rate of change of $f'$ between $2$ and $3$ is $f'_{[2,3]}=\frac{f'(3)-f'(2)}{3-2}= -3$ and the average rate of change of $f'$ between $3$ and $4$ is $f'_{[3,4]}=\frac{f'(4)-f'(3)}{4-3}= -1$. Average these two values, we get

$$ f''(3)\simeq \frac{1}{2}(-3-1) = -2$$

This comes from the definition of tangent line and derivatives as a limit of secant lines, but that's assuming quite a lot on the function $f$. And since $2$ and $4$ are not really "near $3$", that's probably a really bad estimation.

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Study of $f(x)=(x-3)|x-3|+(x-3)+6$: Let $g(x)=x|x|$, so that $f(x)=g(x-3)+(x-3)+6$. For $x>0$, $g(x)=x^2$, so $g'(x)=2x$. Similarly, $g'(x)=-2x$ for $x<0$. For $x=0$, $g'(0)=\lim_{x\to 0}\frac{g(x)-g(0)}{x-0}=\lim_{x\to 0}|x|=0$. To sum up, $g'(x)=2|x|$. It follows that $g''(0)$ does not exist (since $|x|$ has a cusp at $0$).

It also follows that $f(x)$ is differentiable on $\mathbb R$ and $f'(x)=2|x-3|+1$, so $f'(3)=1$ (Chain Rule) abd $f''(x)$ is not differentiable at $3$.

Answer 2

If your $\Delta x$ can't get arbitrarily small you won't get a very accurate derivative. Many functions allow you to get close.

I think you can get close using Fibonacci Numbers. There are at least a couple possible approximations based on the table.

$f'(x)=\lim\limits_{c->0} \frac{f(x+c)-f(x-c)}{2c}$

$f''(x)=\lim\limits_{c->0} \frac{f'(x+c)-f'(x-c)}{2c}$

$f''(x)=\lim\limits_{c->0} \frac{f'(4)-f'(2)}{2}=-2$

$f''(x)=\lim\limits_{c->0}\frac{\frac{f(x+2c)-f(x)}{2c}-\frac{f(x)-f(x-2c)}{2c}}{2c}$

$f''(x)=\lim\limits_{c->0} \frac{f(x+2c)-2f(x)+f(x-2c)}{4c^2}$

All the $\Delta x$ are =1 here. Let c=1.

$f''(3)=\lim\limits_{c->0} \frac{f(5)-2f(3)+f(1)}{4(1)}=\frac{-7}{4}$

So these techniques give you approxmations of -2 and -7/4. Their average would probably be closer to the right value.

Answer 3

This isn't nearly enough information to be sure of the value of $f''(3)$. To take a simple example, $f$ could be the function

$$f(x) = \begin{cases} 3x & \text{for } x < 1.5 \\ 4x - 4 & \text{for } 1.5 \leq x < 2.5 \\ x + 3 & \text{for } 2.5 \leq x < 3.5 \\ -1 & \text{for } 3.5 \leq x < 4.5 \\ 5x - 23 & \text{for } 4.5 \leq x \end{cases}$$

This has $f''(x) = 0$ for all $x$. On the other hand, $f$ could also be

$$f(x) = \begin{cases} 3x & \text{for } x < 1.5 \\ 4x - 4 & \text{for } 1.5 \leq x < 2.5 \\ \frac16x^2 + \frac92 & \text{for } 2.5 \leq x < 3.5 \\ -1 & \text{for } 3.5 \leq x < 4.5 \\ 5x - 23 & \text{for } 4.5 \leq x \end{cases}$$

for which $f''(3) = \frac13$.

The best you can possibly hope for here is an estimation to $f''(3)$. Recall that $f''(3)$ is $\lim_{h \to 0}\frac{f'(3 + h) - f'(3)}{h}$. We only know a few values of $f'$, so we have to use those - it makes sense to use the smallest, $1$ and $-1$. For $h = 1$, $\frac{f'(3 + h) - f'(3)}{h} = -1$. For $h = -1$, $\frac{f'(3 + h) - f'(3)}{h} = -3$. So $-1$ and $-3$ would both be reasonable estimates. A somewhat more reasonable estimate would be the average between them: $\frac{-3 + -1}{2} = -2$.