Derive a polynomial function from a value table

Question

i am stuck with the following excercise:

"The picture shows a value table of a polynomial function f. Of which degree is f? Explain your choice. Determine the function term of f.

$$ \begin{matrix} - & -1 & 0 & 1 & 2 & 3\\ f(x) & -1,66 & 0,75 & 2,333 & 4,083 & 9\\ f'(x) & 2,666 & 2 & 1,333 & 2,666 & 8\\ f''(x) & 0 & -1 & 0 & 3 & 8\\ \end{matrix} $$

I will shortly give you my thoughts so far: I suppose f is of degree 5, because there are two inflection points. So my f(x) should be of form: $$ f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f $$ $$ f'(x) = 5ax^4 + 4bx^3 + 3cx^2 + 2dx + e$$ $$ f''(x) = 20ax^3 + 12bx^2 + 6cx + 2d $$

Now i thought: we have 6 unknowns: a, b, c, d, e, f Now we can choose 6 random equations from the value table to form a linear equation system. For example:

f(-1) = -1,66

f(0) = 0,75

f(1) = 2,333

f(2) = 4,083

f(3) = 9

f'(-1) = 2,666

Unfortunately i obtain solutions that do not deliver a reasonable f. $$\begin{matrix} a & \frac{107}{600000}\\ b & \frac{9923}{120000}\\ c & \frac{73}{120000}\\ d & \frac{59543}{120000}\\ e & \frac{199693}{100000}\\ f & 0,75\\ \end{matrix}$$

Can you guys help me? Best regards, Philipp

Answer 1

Hint:

Note that $$f''(x)=x^2-1$$

Answer 2

We start by examining $f''(x)$. Note the second differences of $f''(x)$ are constant, as shown below:

$\begin{matrix} x & -1 & 0 & 1 & 2 & 3\\ f''(x) & 0 & -1 & 0 & 3 & 8\\ & & -1 & 1 & 3 & 5 \\ & & & 2 & 2 & 2 \end{matrix}$

Therefore, we claim $f''(x)$ can be modeled by a quadratic in the given interval. Setting $f''(x) = ax^2 + bx + c$, we note $f''(0) = c = -1$. Next, we form a system of equations: $f''(1) = a + b - 1 = 0, f''(2) = 4a + 2b - 1 = 3 \implies a = 1, b = 0$. Therefore, we say $f''(x) = x^2 - 1$.

Now, we integrate: $f'(x) = \int {f''(x) dx} = \int {x^2 - 1 dx} = \frac{x^3}{3} - x + C$. We use $f'(0) = 2$ to show that $C = 2$, so $f'(x) = \frac{x^3}{3} - x + 2$.

We again integrate: $f(x) = \int {f'(x) dx} = \int {\frac{x^3}{3} - x + 2 dx} = \frac{x^4}{12} - \frac{x^2}{2} + 2x + C_1$. We use $f(0) = .75$ to show $C_1 = .75$, so $f(x) = \frac{x^4}{12} - \frac{x^2}{2} + 2x + \frac{3}{4}$.

Hope this helps!