Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 07
• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QB 1 • QB 2 • QB 3(a) • QB 3(b) • QB 3(c) • QB 4(a) • QB 4(b) • QB 4(c) • QB 5(a) • QB 5(b) • QB 6(a) • QB 6(b) • QB 7(a) • QB 7(b) • QB 7(c) • QB 8 •
Question A 07 

Let be a function that is continuous at . Which of the following statements is correct? Recall that if there exists any function for which a statement is false then the statement is false. (i) If exists and then is an extremum. (ii) If is an extremum then exists and (iii) If exists for all and , , and then is a minimum. (iv) If , and exist and then is not an extremum. (v) None of the above are true. 
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 

If the derivative at some point is zero and it also changes sign around the point, then the point is an extremum. However, the converse is not true; may not exist, but it can have a local max/min at a singular point. 
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 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. (i) Incorrect. exists and , but the derivative is everywhere nonnegative so is NOT an extremum. (ii) Incorrect. has a minimum at , but the derivative does NOT exist at (a corner). (iii) Incorrect. A function can be decreasing at (1) and increasing at 1, but have a local max at . Consider a function . (iv) Correct. Because if and , this means that is an extremum for . (Applying the second derivative test for ). Now suppose for example which implies that is a minimum point for , and therefore by the definition for any in the neighborhood of 0 we have , which is equivalent to being increasing in the neighborhood of 0, so cannot have an extremum at . By the similar argument, we can get the same result in the case of . 