• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q4 (e) • Q5 • Q6 • Q7 • Q8 •
Question 07 

FullSolution Problems. In questions 28, justify your answers and show all your work. If a box is provided, write your final answer there. Simplification of answers is not required unless explicitly requested. Determine and carefully justify an upper bound for the absolute value of the error that would result if the second degree Maclaurin polynomial was used to estimate , where Your upper bound may be left unsimplified ("Calculatorready") but it must be a specific numerical value. NOTE: You are not required to find and no credit will be given for finding . 
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 

We want to know the value of (or at least an upper bound of)
Taylor's remainder formula says that there is a such that
where M is an upper bound for the absolute value of the third derivative on . 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 1 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. Taylor's remainder formula says that where M is an upper bound for the absolute value of the third derivative on the interval . Computing the third derivative via the product rule gives Notice that is increasing on and thus obtains its maximum at x = 1, i.e. for . The function is a parabola with roots at 0 and 1. Hence its minimum occurs at the midpoint of the roots, namely at . Since the value at the endpoints x = 0 and x = 1 is zero, the minimum value of the parabola maximizes its absolute value. Thus, Since we want to bound the absolute value of the third derivative, we know that
Hence, we have as required. 
Solution 2 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. Proceed as in solution 1 to see that We now need a bound on this function on . Notice that the third derivative is neither monotonically increasing nor is it monotonically decreasing on the interval. Hence we are not guaranteed to find the maximum value by just looking at the endpoints. In fact, at the endpoints we have , which is certainly not the maximum of the absolute value . Instead we do it the proper way: Take the derivative and set it to zero Recall that . Using the quadratic formula, we see that only the positive such root is in . Let
Then, our function obtains an extreme value at . Since the function is nonzero at this point, we know that is maximal at . Thus, we have that completing the solution. 