Q. 14 - Calculus

Question

Let $f\left(x\right)=4x^3-5x^2-6x+7$. Find the first- through fourth-order Maclaurin polynomials, $P_1\left(x\right), P_2\left(x\right), P_3\left(x\right),$ and $P_4\left(x\right)$, for $f$ . Explain why $f\left(x\right)=P_3\left(x\right)=P_4\left(x\right)$. Graph $f\left(x\right), P_1\left(x\right), P_2\left(x\right)$, and $P_3\left(x\right)$.

Step-by-Step Solution

Answer

The Maclaurin polynomials are,

$$\begin{gathered} P_1\left(x\right)=7-6x \\ {P}_2\left(x\right)=7-6x-5x^2 \\ {P}_3\left(x\right)=7-6x-5x^2+4x^3 \\ {P}_4\left(x\right)=7-6x-5x^2+4x^3 \end{gathered}$$

The graph for $f\left(x\right),P_1\left(x\right),P_2\left(x\right),P_3\left(x\right)$is,

Step 1. Given Information.

The functions are,

$f\left(x\right)=4x^3-5x^2-6x+7$.

Step 2. Formula of the Maclaurin polynomials.

The first-, second, third-, and fourth-order Maclaurin polynomials, that is, $P_1\left(x\right), P_2\left(x\right), P_3\left(x\right),$ and $P_4\left(x\right)$ are given by,

$$\begin{gathered} P_1\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x \\ {P}_2\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2 \\ {P}_3\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3 \\ {P}_4\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right)}{4!}{x}^4 \end{gathered}$$

Step 3. Finding all the Maclaurin polynomials.

The value at $x=0$ is,

$$\begin{gathered} f\left(0\right)=4{\left(0\right)}^3-5{\left(0\right)}^2-6\left(0\right)+7 \\ =7 \end{gathered}$$

The derivatives for the function $f\left(x\right)=4x^3-5x^2-6x+7$ are,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d(4x^3-5x^2-6x+7)}{dx} \\ =4\frac{d\left(x^3\right)}{dx}-5\frac{d\left(x^2\right)}{dx}-6\frac{dx}{dx}+\frac{d7}{dx} \\ =12x^2-10x-6 \\ f\text{'}\left(0\right)=12{\left(0\right)}^2-10\left(0\right)-6 \\ =6 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(12x^2-10x-6)}{dx} \\ =12\frac{d\left(x^2\right)}{dx}-10\frac{dx}{dx}-\frac{d6}{dx} \\ =24x-10 \\ f\text{'}\text{'}\left(0\right)=24\left(0\right)-10 \\ =-10 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d(24x-10)}{dx} \\ =24\frac{dx}{dx}-\frac{d10}{dx} \\ =24 \\ f\text{'}\text{'}\text{'}\left(0\right)=24 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(0\right)=\frac{d24}{dx} \\ =0 \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=0 \end{gathered}$$

The first-, second-, third-, and fourth-, order polynomials are,

$$\begin{gathered} P_1\left(x\right)=7-6x \\ {P}_2\left(x\right)=7-6x-5x^2 \\ {P}_3\left(x\right)=7-6x-5x^2+4x^3 \\ {P}_4\left(x\right)=7-6x-5x^2+4x^3 \end{gathered}$$

Step 4. Explanation and graph.

Here, $P_3\left(x\right)=P_4\left(x\right)$. This is because for any polynomial function $f$ of degree $n$, the $nth$ Maclaurin polynomial for $f$, $P_m\left(x\right)=f\left(x\right) where m\ge n$.

The graph of $f\left(x\right),P_1\left(x\right),P_2\left(x\right),P_3\left(x\right)$ are: