Question

find the taylor polynomials of orders 0, 1, 2, and 3 generated by f at a
$f(x)=e^{4x},a = 0$
$p_{0}(x)=1$(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 \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). Here \(a = 0\), so \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\).
First, find the derivatives of \(f(x)=e^{4x}\).
The first - derivative \(f^{\prime}(x)=4e^{4x}\), the second - derivative \(f^{\prime\prime}(x)=16e^{4x}\), the third - derivative \(f^{(3)}(x)=64e^{4x}\).

Step2: Calculate \(P_1(x)\)

For \(n = 1\), \(P_1(x)=\frac{f(0)}{0!}x^0+\frac{f^{\prime}(0)}{1!}x^1\).
Since \(f(0)=e^{4\times0}=1\) and \(f^{\prime}(0)=4e^{4\times0}=4\).
Then \(P_1(x)=\frac{1}{1}\times1+\frac{4}{1}x=1 + 4x\).

Answer:

\(P_1(x)=1 + 4x\)