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)=\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 \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\).
For \(n = 2\), we need \(f(0)\), \(f^{\prime}(0)\), \(f^{\prime\prime}(0)\).
Given \(f(x)=e^{6x}\), then \(f^{\prime}(x)=6e^{6x}\), \(f^{\prime\prime}(x)=36e^{6x}\).

Step2: Evaluate the function and its derivatives at \(a = 0\)

\(f(0)=e^{6\times0}=1\), \(f^{\prime}(0)=6e^{6\times0}=6\), \(f^{\prime\prime}(0)=36e^{6\times0}=36\).

Step3: Construct \(P_2(x)\)

Using the formula \(P_2(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1+\frac{f^{\prime\prime}(0)}{2!}x^2\).
Since \(0!=1\), \(1!=1\), \(2!=2\), we have \(P_2(x)=1 + 6x+\frac{36}{2}x^2\).

Answer:

\(P_2(x)=1 + 6x+18x^2\)