Question

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

Short answer

The Taylor-polynomials are,

$$\begin{gathered} P_1\left(x\right)=6+4(x-1) \\ {P}_2\left(x\right)=6+4(x-1)+3{(x-1)}^2 \\ {P}_3\left(x\right)=6+4(x-1)+3{(x-1)}^2 \end{gathered}$$

Step-by-step solution

Step 1. Given Information.

The function is,

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

Step 2. Formulae for Taylor polynomials.

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

The first-, second, and third-, order Taylor polynomials at $x=1$, that is, $P_1\left(x\right),P_2\left(x\right),P_3\left(x\right)$are given by,

role="math" localid="1649589044105" $$\begin{gathered} P_1\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)(x-1) \\ {P}_2\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)(x-1)+\frac{f\text{'}\text{'}\left(0\right)}{2!}{(x-1)}^2{P}_3\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)(x-1)+\frac{f\text{'}\text{'}\left(0\right)}{2!}{(x-1)}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{(x-1)}^3 \end{gathered}$$

Step 3. Finding the Taylor polynomials.

The value at $x=1$is,

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

Finding the derivatives of the function $f\left(x\right)=3x^2-2x+5$are,

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

Also,

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

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(6\right)}{dx} \\ =0 \\ f\text{'}\text{'}\text{'}\left(1\right)=0 \end{gathered}$$

Hence, the Taylor polynomials are,

role="math" localid="1649589449293" $$\begin{gathered} P_1\left(x\right)=6+4(x-1) \\ {P}_2\left(x\right)=6+4(x-1)+3{(x-1)}^2 \\ {P}_3\left(x\right)=6+4(x-1)+3{(x-1)}^2 \end{gathered}$$

Step 4. Explanation.

Here $P_2\left(x\right)=P_3\left(x\right)$. This is because for any polynomial function $f$of degree $n$the $mth$Taylor polynomial for$f$,$P_m\left(x\right)=f\left(x\right)whenm\ge n$.