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 the Taylor polynomial formula

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(a\) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). Here \(a = 0\), so \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\), and \(f(x)=e^{6x}\).
First, find the derivatives of \(f(x)\):

• \(f(x)=e^{6x}\), then \(f(0)=e^{0}=1\)
• \(f^{\prime}(x)=6e^{6x}\), so \(f^{\prime}(0)=6e^{0}=6\)
• \(f^{\prime\prime}(x)=6^{2}e^{6x}\), so \(f^{\prime\prime}(0)=6^{2}e^{0}=36\)
• \(f^{(3)}(x)=6^{3}e^{6x}\), so \(f^{(3)}(0)=6^{3}e^{0}=216\)

Step2: Calculate \(P_3(x)\)

Using the formula \(P_3(x)=\frac{f(0)}{0!}x^{0}+\frac{f^{\prime}(0)}{1!}x^{1}+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}\)

• \(\frac{f(0)}{0!}x^{0}=\frac{1}{1}\times1 = 1\)
• \(\frac{f^{\prime}(0)}{1!}x^{1}=\frac{6}{1}x=6x\)
• \(\frac{f^{\prime\prime}(0)}{2!}x^{2}=\frac{36}{2}x^{2}=18x^{2}\)
• \(\frac{f^{(3)}(0)}{3!}x^{3}=\frac{216}{6}x^{3}=36x^{3}\)

Answer:

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