Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 07

MATH102 December 2016

  •  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  •  

   Other MATH102 Exams

  •  December 2014  •  December 2011  •  December 2012  •  December 2013  •  December 2016  •  December 2015  •  

Question A 07

Let $f(x)$ be a function that is continuous at $x=0$. 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 $f'(0)$ exists and $f'(0)=0$ then $x=0$ is an extremum. (ii) If $x=0$ is an extremum then $f'(0)$ exists and $f'(0)=0.$ (iii) If $f'(x)$ exists for all $x$ and $f'(0)=0$, $f'(1)>0$, and $f'(-1)<0$ then $x=0$ is a minimum. (iv) If $f'(0)$, $f''(0)$ and $f'''(0)$ exist and $f'(0)=f''(0)=0\neq f'''(0)$ then $x=0$ 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; $f'(x_{0})$ 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. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. $f(x)=x^{3},\ f'(x)=3x^{2},\ f'(0)$ exists and $f'(0)=0$, but the derivative is everywhere non-negative so $x=0$ is NOT an extremum. (ii) Incorrect. $f(x)=|x|$ has a minimum at $x=0$, but the derivative does NOT exist at $x=0$ (a corner). (iii) Incorrect. A function can be decreasing at (-1) and increasing at 1, but have a local max at $x=0$. Consider a function $f'(x)=x(x-{\frac {1}{2}})(x+{\frac {1}{2}})$. (iv) Correct. Because if $f''(0)=0$ and $f'''(0)\neq 0$, this means that $(0,0)$ is an extremum for $f'$. (Applying the second derivative test for $f'$). Now suppose for example $f'''(0)>0$ which implies that $(0,0)$ is a minimum point for $f'$, and therefore by the definition for any $x$ in the neighborhood of 0 we have $f'(x)>0$, which is equivalent to $f$ being increasing in the neighborhood of 0, so $f$ cannot have an extremum at $x=0$ . By the similar argument, we can get the same result in the case of $f'''(0)<0$.