Science:Math Exam Resources/Courses/MATH100/December 2011/Question 04 (a)

MATH100 December 2011

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Question 04 (a)
Full-Solution Problems. In questions 2-8, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly stated. Let T2(x) be the second degree Taylor polynomial about a = 8 for $f(x)={\sqrt[{3}]{x}}.$ Find T2(x) and simplify your answer.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that the general forumla for a second degree Taylor polynomial is: $T_{2}(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}$

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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. The general forumla for a second degree Taylor polynomial is: $T_{2}(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}$ To find this for ƒ(x), we need to first find ƒ'(x) and ƒ″(x): $f'(x)=({\sqrt[{3}]{x}})'=(x^{1/3})'={\frac {1}{3}}x^{-2/3}$ $f''(x)=({\frac {1}{3}}x^{-2/3})'=-{\frac {2}{9}}x^{-5/3}$ And now we can evaluate at a = 8 to get $f(8)=8^{1/3}=2$ $f'(8)={\frac {1}{3}}8^{-2/3}={\frac {1}{3}}\cdot {\frac {1}{4}}={\frac {1}{12}}$ $f''(8)=-{\frac {2}{9}}8^{-5/3}=-{\frac {2}{9}}\cdot {\frac {1}{2^{5}}}=-{\frac {2}{9\cdot 32}}=-{\frac {1}{144}}$ We can now plug these in to the equation for the second Taylor polynomial and obtain: $T_{2}(x)=2+{\frac {1}{12}}(x-8)-{\frac {1}{288}}(x-8)^{2}$ (This is simplified enough for the purpose of Taylor polynomials).

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