Question

find the taylor polynomials of orders 0, 1, 2, and 3 generated by f at a.
f(x)=e^{6x}, a = 0

p_0(x)=1 (simplify your answer.)
p_1(x)=1 + 6x
p_2(x)=1 + 6x + 18x^2
p_3(x)=\square

Explanation:

Step1: Recall Taylor polynomial formula

The Taylor polynomial of order \(n\) for \(f(x)\) about \(a\) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). Here \(a = 0\), so \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\).
First, find derivatives of \(f(x)=e^{6x}\).
\(f(x)=e^{6x}\), then \(f(0)=e^{0}=1\)
\(f^{\prime}(x)=6e^{6x}\), \(f^{\prime}(0)=6\)
\(f^{\prime\prime}(x)=6^{2}e^{6x}\), \(f^{\prime\prime}(0)=6^{2}\)
\(f^{\prime\prime\prime}(x)=6^{3}e^{6x}\), \(f^{\prime\prime\prime}(0)=6^{3}\)

Step2: Calculate \(P_3(x)\)

For \(n = 3\), \(P_3(x)=\frac{f(0)}{0!}x^{0}+\frac{f^{\prime}(0)}{1!}x^{1}+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{\prime\prime\prime}(0)}{3!}x^{3}\)
Substitute \(f(0) = 1\), \(f^{\prime}(0)=6\), \(f^{\prime\prime}(0)=36\), \(f^{\prime\prime\prime}(0)=216\)
\(P_3(x)=\frac{1}{1}\times1+\frac{6}{1}x+\frac{36}{2}x^{2}+\frac{216}{6}x^{3}\)
Simplify each term:
\(\frac{36}{2}=18\), \(\frac{216}{6} = 36\)

Answer:

\(P_3(x)=1 + 6x+18x^{2}+36x^{3}\)