Science:Math Exam Resources/Courses/MATH104/December 2013/Question 06 (b)

MATH104 December 2013

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Question 06 (b)
Use the formuala $|{\text{Error}}|\leq {\frac {M}{2}}(x-a)^{2},$ where $M>0$, to estimate an error bound for your approximation of ${\sqrt[{3}]{7}}$ from part (a). It will be useful to remember how to find such an $M.$

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
To find a value for $M$, it would suffice to find an upper bound on the absolute value of the second derivative and use this as the value for $M$.

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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. Our function we are considering is $\displaystyle f(x)=x^{1/3}$. Taking derivatives gives ${\begin{aligned}f'(x)&={\frac {x^{-2/3}}{3}}\\f''(x)&={\frac {-2x^{-5/3}}{9}}\end{aligned}}$ Since the linear approximation is centred about $a=8$ we are looking for the maximum value of $|f''(x)|={\frac {2}{9}}x^{-5/3}$ on the interval $\displaystyle {}[7,8]$. Since $x^{-5/3}$ is decreasing, the maximum value is attained at $x=7$, that is, $M=|f''(7)|.$ Overall we find that $|\mathrm {Error} |\leq {\frac {1}{2}}M(x-a)^{2}={\frac {1}{2}}\cdot {\frac {2}{9}}7^{-5/3}(7-8)^{2}={\color {blue}{\frac {7^{-5/3}}{9}}.}$

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