Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 13 (d)

MATH 180 December 2017

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Question 13 (d)
(d) Is ${\textstyle Q(2.8)>f(2.8)}$, ${\textstyle Q(2.8)<f(2.8)}$ or ${\textstyle Q(2.8)=f(2.8)}$? Determine which of these three options is correct, and justify 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
Use the quadratic approximation with a cubic remainder.

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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 the Hint, we consider the quadratic approximation with cubic remainder and write $$f(x)=Q(x)+{\dfrac {f'''(c)}{3!}}(x-1)^{3},$$ where ${\textstyle c}$ is a number between ${\textstyle x}$ and ${\textstyle 1}$. For ${\textstyle x=2.8}$, we have $$f(2.8)=Q(2.8)+{\dfrac {f'''(c)}{6}}(2.8-1)^{3},$$ where ${\textstyle c}$ is a number between ${\textstyle 2.8}$ and ${\textstyle 1}$. Let us find the third derivative of ${\textstyle f}$. Recalling that ${\textstyle f''(x)=-{\dfrac {1}{x^{2}}}=-x^{-2}}$, we have ${\textstyle f'''(x)=-(-2)x^{-3}=2x^{-3}={\dfrac {2}{x^{3}}}}$. For ${\textstyle 1<c<2.8}$, $$f'''(c)={\dfrac {2}{c^{3}}}>0,$$ because ${\textstyle c>0}$ in particular. Putting everything together, we see that $$f(2.8)=Q(2.8)+{\dfrac {f'''(c)}{6}}1.8^{3},$$ with ${\textstyle f'''(c)>0}$. Therefore the last term is positive and we conclude that $$f(2.8)>Q(2.8),$$ or $$Q(2.8)<f(2.8).$$ Answer: we have ${\textstyle {\color {blue}Q(2.8)<f(2.8)}}$.