Q. 4 - Calculus

Question

Suppose f is ant cubic polynomial function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$ prove that coefficients of f a, b, c, d can be expressed in terms of values of f(x) and its derivatives at the point x=2 

Step-by-Step Solution

Answer

We give the values of a, ,b, c, d in terms of function and its derivative at x=2

Step 1: Given information

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

Step 2: We find all the necessary derivatives

We get,

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

Compute the derivative at x=2

We get,

$$\begin{gathered} f\left(2\right)=8a+4b+2c+d \left(1\right) \\ f\text{'}\left(2\right)=12a+4b+c \left(2\right) \\ f"\left(2\right)=12a+2b \left(3\right) \\ f"\text{'}\left(2\right)=6a \left(4\right) \end{gathered}$$

Step 3: Write in terms of the function and its derivative at x=2

We get, 

Substitute 4 in 3

$$\begin{gathered} f"\left(2\right)=2f"\text{'}\left(2\right)+2b \\ b=\frac{f"\left(2\right)-2f"\text{'}\left(2\right)}{2} \left(5\right) \\ a=\frac{f"\text{'}\left(2\right)}{2} \left(6\right) \end{gathered}$$

Substituting 5 and 6 in 2 we get

$c=2f"\text{'}\text{'}\left(2\right)-2f"\left(2\right)+f\text{'}\left(2\right)$

Substituting all values in 1 we get\

$a=f\left(2\right)-2f\text{'}\left(2\right)+2f\text{'}"\left(2\right)-\frac{8}{6}f"\text{'}\left(2\right)$