Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08

MATH100 December 2018

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Question 08
Let $f(x)$ be a differentiable function satisfying the following properties: $f(0)=1,f'(0)=-2,f''(0)=3;$ and $f'''(x)={\frac {6-x^{2}}{9-\cos(x)}}$ Determine with proof an approximation $A$ to $f(2)$ with the property that the error in the approximation is at most $1$ .

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
Use the Maclaurin polynomial.

Hint 2
Use the Lagrange Remainder Theorem to estimate the error of your approximation.

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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. We know that $f$ is at least three times differentiable, so we will use the second degree Maclaurin polynomial, which is $A=T_{2}(x)=f(0)+f'(0)x+{\frac {f''(0)}{2}}\cdot x^{2}=1-2x+{\frac {3x^{2}}{2}}.$ That is, $f(2)$ is approximated by $\color {blue}T(2)=1-2\cdot 2+{\frac {3\cdot 2^{2}}{2}}=3$ . To estimate the error of our approximation, by the Lagrange Remainder Theorem we have $f(2)-T_{2}(2)={\frac {f'''(c)}{3!}}\cdot (2-0)^{3}={\frac {\frac {6-c^{2}}{9-\cos(c)}}{6}}\cdot 8={\frac {4(6-c^{2})}{3(9-\cos(c))}},$ where $c$ is in the interval $[0,2]$ . Note that this expression is always positive for $c$ in $[0,2]$ . Furthermore, the numerator is decreasing on this interval and the denominator attains its minimum at $c=0$ , which implies that this error is maximised at $c=0$ . We conclude that the error $\color {blue}\|f(2)-T_{2}(2)\|\leq {\frac {4\cdot 6}{3\cdot 8}}=1$ , as required.