Science:Math Exam Resources/Courses/MATH100/December 2015/Question 03 (iv)

MATH100 December 2015

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Question 03 (iv)
Let $T_{3}(x)=24+6(x-3)+12(x-3)^{2}+4(x-3)^{3}$ be the third-degree Taylor polynomial for some function $f(x)$, expanded about $a=3$. What is $f''(3)$? 2 3 4 6 12 24

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
Try to write the general formula for the third-degree Taylor polynomial for $f(x)$ about $a=3$ and compare that with the given formula.

Hint 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 third-degree Taylor polynomial for $f(x)$, expanded about $a=3$ is $T_{3}(x)=f(3)+f'(3)(x-3)+{\frac {\color {RedOrange}f''(3)}{2!}}(x-3)^{2}+{\frac {f'''(3)}{3!}}(x-3)^{3},$ Comparing this with the formula given in the question we get $f(3)+f'(3)(x-3)+{\color {OliveGreen}{\frac {f''(3)}{2!}}}(x-3)^{2}+{\frac {f'''(3)}{3!}}(x-3)^{3}=24+6(x-3)+{\color {OliveGreen}12}(x-3)^{2}+4(x-3)^{3}.$ Comparing the coefficient of $(x-3)^{2}$, we get ${\frac {\color {RedOrange}f''(3)}{2!}}=12,$ and hence ${\color {RedOrange}f''(3)}=12\times (2!)=12\times 2={\color {RedOrange}24}.$ Therefore, the answer is F.