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.)

$p_1(x)=\square$

Explanation:

Step1: Recall the Taylor polynomial formula

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(a\) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). When \(a = 0\), it becomes \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\).

Step2: Find the first - derivative of \(f(x)\)

Given \(f(x)=e^{6x}\), using the chain rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) with \(u = 6x\), we have \(f^\prime(x)=6e^{6x}\).

Step3: Evaluate \(f(x)\) and \(f^\prime(x)\) at \(x = 0\)

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

Step4: Calculate \(P_1(x)\)

Using the Taylor polynomial formula \(P_1(x)=\frac{f(0)}{0!}x^0+\frac{f^\prime(0)}{1!}x^1\). Since \(0!=1\) and \(1!=1\), \(P_1(x)=1 + 6x\).

Answer:

\(P_1(x)=1 + 6x\)