MATH110 April 2018
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q2 (e) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q7 (c) • Q7 (d) • Q7 (e) • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q9 • Q10 •
Question 02 (e)
Find the absolute minimum of .
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!
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.
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies.
By the Hint, the critical points are candidates for the points at which an extremum occurs. Recall that a function has a critical number
in the domain if it satisfies one the the following conditions.
does not exist
a is an end point.
So, we find critical points of .
Observe that has a domain (since the logarithmic function can only take positive arguments), and on the domain the given function is differentiable everywhere. In other words, there's no point at which doesn't exist. Also, we can see that our domain doesn't have end point because it is an open interval .
Now, we find the roots of the first derivative of . The first derivative of is given by
Setting this to zero and rearranging gives By multiplying through by , we obtain .
Therefore, is the only critical point of .
It remains to check that this is indeed a global minimum. Indeed, it suffices to show that for a single (in ), one has . If , then as required.
Answer: The function has a global minimum at .
Click here for similar questions
MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, Pages using DynamicPageList parser function, Pages using DynamicPageList parser tag