Q. 13 - Calculus

Question

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

Step-by-Step Solution

Answer

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

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

Step 1. Given Information.

The function is,

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

Step 2. The formula for first, second, and third-order Maclaurin polynomials.

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

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

Step 3. Finding the first second and third-order Maclaurin polynomials.

Finding the value at $x=0$,

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

Finding the derivatives of the function,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d(3x^2-2x+5)}{dx} \\ =3\frac{d\left(x^2\right)}{dx}-2\frac{dx}{dx}+5\frac{d0}{dx} \\ =6x-2 \\ f\text{'}\left(0\right)=6\left(0\right)-2 \\ =-2 \end{gathered}$$

Also,

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

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d6}{dx} \\ =0 \\ f\text{'}\text{'}\text{'}\left(0\right)=0 \end{gathered}$$

Therefore, the first, second and third-order Maclaurin polynomials are,

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

Step 4. Explanation and Graph.

Here,$P_2\left(x\right)=P_3\left(x\right)$. This is because for any polynomial function $f$ of degree $n$, the $nth$ Maclaurin Polynomial is, $P_m\left(x\right)=f\left(x\right) where m\ge n$.

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