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 14 (a)
Let be a function so that
- , exist and are continuous for all , and
- for all .
Show that has at least one zero in the open interval .
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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.
Use the intermediate value theorem.
For any number , can be expanded as .
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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To find a zero of , we use the intermediate value theorem. Note that by the first condition, is continuous.
To this end, it is enough to find two values of at which has a negative and positive function value, respectively.
Expanding the inequality in the second condition, we get
Since and plugging these numbers into the inequalities, we get
Therefore, by the intermediate value theorem, has at least one zero in the interval , which is contained in the interval .