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)=\square (simplify your answer.)

Explanation:

Step1: Recall the formula for Taylor polynomial of order 0

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(a\) is given by \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). For \(n = 0\), \(P_0(x)=\frac{f(a)}{0!}(x - a)^0\).
Since \(a = 0\) and \(f(x)=e^{6x}\), then \(f(0)=e^{6\times0}\).

Step2: Calculate \(f(0)\)

We know that for any real number \(t\), \(e^{t}\) has the property that \(e^{0}=1\). So when \(t = 6\times0=0\), \(f(0)=e^{0}=1\). And \((x - 0)^0 = 1\) (by the rule \(x^0=1,x
eq0\), and here when considering the polynomial, we take the value as \(1\) for all \(x\) in the domain of the polynomial). Also, \(0!=1\).

Answer:

\(1\)