Question

Prove that if f is any cubic polynomial function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$then the coefficients of f are completely determined by the values of f(x) and its derivative at x=0 as follows

$a=\frac{f"\text{'}\left(0\right)}{6};b=\frac{f"\left(0\right)}{2};c=f\text{'}\left(0\right);d=f\left(0\right)$

Short answer

We determine the values of a, b, c, d in terms of f and its derivatives at x=0

Step-by-step solution

Given information

We are given a cubic polynomial$f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Find the derivatives of the function

We have,

$$\begin{gathered} f\left(x\right)=a{x}^3+b{x}^2+cx+d \\ f\text{'}\left(x\right)=3a{x}^2+2bx+c \\ f"\left(x\right)=6ax+2b \\ f"\text{'}\left(x\right)=6a \end{gathered}$$

Now compute all the values at x=0

We get,

$$\begin{gathered} f\left(0\right)=d\left(1\right) \\ f\text{'}\left(0\right)=c\left(2\right) \\ f"\left(0\right)=2b \\ b=\frac{f"\left(0\right)}{2}\left(3\right) \\ f"\text{'}\left(0\right)=6a \\ a=\frac{f"\text{'}\left(0\right)}{6}\left(4\right) \end{gathered}$$

We determine the values of a, b, c, d in terms of f and its derivatives at x=0