Q. 9 - Calculus

Question

If $f\left(x\right)=4x^3-5x^2+6x+1 and P_3\left(x\right)$ is the third Taylor polynomial for f at −1, what is the third remainder $R_3\left(x\right)$? What is $R_4\left(x\right)$? (Hint: You can answer this question without finding any derivatives.) 

Step-by-Step Solution

Answer

The required values are $R_3\left(x\right)=0 and R_4\left(x\right)=0$

Step 1. Given Information

The given function is $f\left(x\right)=4x^3-5x^2+6x+1$

Step 2. Calculation

The formula to calculate the remainder is $R_n\left(x\right)=\frac{f^{n+1}\left(c\right)}{(n+1)!}{(x-x_0)}^{n+1}$

Substitute n as 3 to find the third remainder of the function.

$$\begin{gathered} R_3\left(x\right)=\frac{f^{3+1}\left(c\right)}{(3+1)!}{(x-x_0)}^{3+1} \\ =\frac{f^4\left(c\right)}{4!}{(x-x_0)}^4 \end{gathered}$$

Since the given degree has highest degree 3 which implies that $f^4\left(c\right)=0$

Hence, $R_3\left(x\right)=0$

Similarly,

$$\begin{gathered} R_4\left(x\right)=\frac{f^{4+1}\left(c\right)}{(4+1)!}{(x-x_0)}^{4+1} \\ =\frac{f^5\left(c\right)}{5!}{(x-x_0)}^5 \\ =\frac{0}{5!}{(x-x_0)}^5 \\ =0 \end{gathered}$$