Question

find the taylor polynomial of order 3 generated by f at a
$f(x)=x^{2},a = 4$
a $p_{3}(x)=1 + 8(x - 4)+12(x - 4)^{2}+16(x - 4)^{3}$
b $p_{3}(x)=16 + 8(x - 4)+12(x - 4)^{2}+16(x - 4)^{3}$
c $p_{3}(x)=16 + 8(x - 4)+(x - 4)^{2}$
d $p_{3}(x)=1 + 32(x - 4)+192(x - 4)^{2}+1,024(x - 4)^{3}$

Explanation:

Step1: Recall the Taylor polynomial formula

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(a\) is \(P_{n}(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}\), where \(f^{(k)}(x)\) is the \(k\) - th derivative of \(f(x)\).

For \(n = 3\), \(P_{3}(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^{2}+\frac{f^{(3)}(a)}{3!}(x - a)^{3}\)

Step2: Calculate the function and its derivatives

Given \(f(x)=x^{2}\), then:

• \(f(a)\): When \(a = 4\), \(f(4)=4^{2}=16\)
• \(f^{\prime}(x)\): Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(f^{\prime}(x)=2x\), so \(f^{\prime}(4)=2\times4 = 8\)
• \(f^{\prime\prime}(x)\): \(f^{\prime\prime}(x)=2\), so \(f^{\prime\prime}(4)=2\)
• \(f^{(3)}(x)\): \(f^{(3)}(x)=0\)

Step3: Substitute into the Taylor polynomial formula

\(P_{3}(x)=f(4)+f^{\prime}(4)(x - 4)+\frac{f^{\prime\prime}(4)}{2!}(x - 4)^{2}+\frac{f^{(3)}(4)}{3!}(x - 4)^{3}\)
Since \(\frac{f^{\prime\prime}(4)}{2!}=\frac{2}{2}=1\) and \(\frac{f^{(3)}(4)}{3!}=0\)
\(P_{3}(x)=16 + 8(x - 4)+(x - 4)^{2}\)

Answer:

C. \(P_{3}(x)=16 + 8(x - 4)+(x - 4)^{2}\)