Question

Let $f\left(x\right)=4x^3-5x^2-6x+7$. Find the first- through fourth-order Taylor polynomials, $P_1\left(x\right),P_2\left(x\right),P_3\left(x\right),$and $P_4\left(x\right)$, for $f$at $1$. Explain why $f\left(x\right)=P_3\left(x\right)=P_4\left(x\right)$.

Short answer

The Taylor polynomials are,

$$\begin{gathered} P_1\left(x\right)=-4x+4 \\ {P}_2\left(x\right)=7x^2-18x+11 \\ {P}_3\left(x\right)=4x^3-5x^2-6x+7 \\ {P}_4\left(x\right)=4x^3-5x^2-6x+7 \end{gathered}$$

Step-by-step solution

Step 1. Given Information.

The function is,

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

Step 2. Formulae for Taylor polynomials.

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

role="math" localid="1649590527241" $$\begin{gathered} P_1\left(x\right)=f\left(1\right)+f\text{'}\left(1\right)(x-1) \\ {P}_2\left(x\right)=f\left(1\right)+f\text{'}\left(1\right)(x-1)+\frac{f\text{'}\text{'}\left(1\right)}{2!}{(x-1)}^2 \\ {P}_3\left(x\right)=f\left(1\right)+f\text{'}\left(1\right)(x-1)+\frac{f\text{'}\text{'}\left(1\right)}{2!}{(x-1)}^2+\frac{f\text{'}\text{'}\text{'}\left(1\right)}{3!}{(x-1)}^3 \\ {P}_4\left(x\right)=f\left(1\right)+f\text{'}\left(1\right)(x-1)+\frac{f\text{'}\text{'}\left(1\right)}{2!}{(x-1)}^2+\frac{f\text{'}\text{'}\text{'}\left(1\right)}{3!}{(x-1)}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(x\right)}{4!}{(x-1)}^4 \end{gathered}$$

Step 3. Finding the Taylor polynomials.

The value at $x=1$is,

$$\begin{gathered} f\left(1\right)=4{\left(1\right)}^3-5{\left(1\right)}^2-6\left(1\right)+7 \\ =4-5-6+7 \\ =0 \end{gathered}$$

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

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d(4x^3-5x^2-6x+7)}{dx} \\ =4\frac{d\left(x^3\right)}{dx}-5\frac{d\left(x^2\right)}{dx}-6\frac{dx}{dx} \\ =12x^2-10x-6 \\ f\text{'}\left(1\right)=12{\left(1\right)}^2-10\left(1\right)-6 \\ =12-10-6 \\ =-4 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(12x^2-10x-6)}{dx} \\ =12\frac{d\left(x^2\right)}{dx}-10\frac{dx}{dx} \\ =24x-10 \\ f\text{'}\text{'}\left(1\right)=24\left(1\right)-10 \\ =24-10 \\ =14 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d(24x-10)}{dx} \\ =24\frac{dx}{dx} \\ =24 \\ f\text{'}\text{'}\text{'}\left(1\right)=24 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d24}{dx} \\ =0 \\ f\text{'}\text{'}\text{'}\text{'}\left(1\right)=0 \end{gathered}$$

Hence, the Taylor polynomials are,

$$\begin{gathered} P_1\left(x\right)=0-4(x-1) \\ =-4x+4 \\ {P}_2\left(x\right)=0-4(x-1)+\frac{14}{2}{(x-1)}^2=7x^2-18x+11 \\ {P}_3\left(x\right)=0-4(x-1)+\frac{14}{2}{(x-1)}^2+\frac{24}{6}{(x-1)}^3=4x^3-5x^2-6x+7 \\ {P}_4\left(x\right)=0-4(x-1)+\frac{14}{2}{(x-1)}^2+\frac{24}{6}{(x-1)}^3+\frac{0}{24}{(x-1)}^4=4x^3-5x^2-6x+7 \end{gathered}$$

Step 4. Explanation.

Here, $P_3\left(x\right)=P_4\left(x\right)$. This is because for any polynomial function $f$of degree$n$, the$mth$Taylor polynomial for$f$,$P_m\left(x\right)=f\left(x\right)whenm\ge n$.