• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 • Q6 • Q7 • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q9 • Q10 (a) • Q10 (b) • Q11 •
Question 10 (b) 

FullSolution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise. Using the Lagrange Remainder Formula for prove that . 
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 obtained from part (a) and the Lagrange remainder formula to calculate the error. 
Hint 2 

Recall that the Lagrange Remainder formula gives that the remainder (error) in the kth degree Taylor polynomial of f at a is: where is some (undetermined) number in the interval .
for some in the interval . 
Hint 3 

To apply the Lagrange Remainder formula, we must find an such that . 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

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. To apply the Lagrange Remainder Formula, we first need to find an such that , that is, . Since is an even function, we choose the positive root without loss of generality.
where and denote the minimum and maximum values of on the interval , respectively.
for some in in the interval . To find the error bounds, we maximize this function as a function of . But observe that is monotonically decreasing on the interval (the denominator is increasing while the numerator is decreasing (note the minus sign), so overall is decreasing). Hence the minimum and maximum values of the error are attained at and , respectively.
