Science:Math Exam Resources/Courses/MATH100/December 2013/Question 04 (c)

MATH100 December 2013

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Question 04 (c)
Short-Answer Question. Simplify your answer as much as possible and show your work. Let $\displaystyle f(x)=4x^{3}$. What is the 2nd degree Taylor polynomial $\displaystyle T_{2}(x)$ centred at a = 1 for $\displaystyle f(x)$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that the $n$-th degree Taylor polynomial $T_{n}(x)$ centred at $x=a$ for a function $f(x)$ is given by $T_{n}(x)=\sum _{i=0}^{n}c_{i}(x-a)^{i}$ , where $c_{i}$ is a coefficient.

Hint 2
Recall also that the coefficients are given by $c_{i}={\frac {f^{(i)}(a)}{i!}}$, i.e., $T_{n}(x)=\sum _{i=0}^{n}{\frac {f^{(i)}(a)}{i!}}(x-a)^{i}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By definition of the Taylor polynomials, ${\begin{aligned}T_{2}(x)&=\sum _{i=0}^{2}{\frac {f^{(i)}(a)}{i!}}(x-a)^{i}\\&={\frac {f^{(0)}(a)}{0!}}(x-a)^{0}+{\frac {f^{(1)}(a)}{1!}}(x-a)^{1}+{\frac {f^{(2)}(a)}{2!}}(x-a)^{2}\\&=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}\end{aligned}}$ Computing the values of the derivatives of $f$ at 1, we find that $f(1)=4,\,f'(1)=12,\,f''(1)=24$. Therefore ${\begin{aligned}T_{2}(x)&=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}\\&={\color {blue}4+12(x-1)+12(x-1)^{2}}\\&={\color {blue}12x^{2}-12x+4}\end{aligned}}$

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