MATH100 December 2016
• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 • Q9 (a) (i) • Q9 (a) (ii) • Q9 (a) (iii) • Q9 (b) (i) • Q9 (b) (ii) • Q9 (b) (iii) • Q9 (c) (i) • Q9 (c) (ii) • Q9 (c) (iii) • Q10 (a) • Q10 (b) • Q11 (a) • Q11 (b) • Q12 • Q13 (a) • Q13 (b) • Q14 (a) • Q14 (b) • Q2 (b) •
Question 07 (a)
Let be a continuous function on the open interval . Which of the following four statements are always true?
Select all that apply.
If and , then there is at least one zero of inside .
If has a local minimum at in , then .
must have both maximum and minimum inside .
If for some point in , then has a local minimum at .
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!
Consider the intermediate value theorem.
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(A) True — by the intermediate value theorem.
(B) False — the derivative of a function may not even exist at a local minimum! For example, consider on Although has its local minimum at does not exist.
(C) False — just take the function on which has no minimum or maximum on
(D) False — the derivative could be everywhere positive. For example, consider on While for any point in , it doesn't have local minimum on this interval. This is because is strictly increasing on the interval, i.e.,