Science:Math Exam Resources/Courses/MATH110/December 2010/Question 03 (c)

MATH110 December 2010

• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q6 (c) • Q7 • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) • Q9 (e) • Q9 (f) • Q10 •

Other MATH110 Exams

• December 2011 • December 2010 • April 2012 • April 2011 • December 2012 • April 2013 • December 2013 • April 2014 • December 2014 • December 2015 • April 2017 • April 2016 • December 2016 • December 2017 • April 2018 •

[hide] Question 03 (c)
Give an example of a function $\displaystyle f$ defined in the interval $[-1,1]$ satisfying $\displaystyle f(-1)=-1$ and $\displaystyle f(1)=1$ , such that $f(x)\not =0$ for any $x$ in $[-1,1]$ .

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!

[show] Hint
The function you draw should have a discontinuity somewhere on its domain.

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.

[show] Solution
The simplest example of such a function would be the piecewise function: $f(x)={\begin{cases}-1&x<0\\1&x\geq 0\end{cases}}$ The function $f$ is clearly defined on $[-1,1]$ , but jumps from -1 to 1 at x = 0. Thus at no point in $[-1,1]$ does $f(x)=0$ . Two other variations: $f(x)={\begin{cases}1/x&x\not =0\\1&x=0\end{cases}}$ In this case, the function $1/x$ is never equal to zero; however, we must add the point $(0,1)$ in the piecewise function to make it defined everywhere in our domain. $f(x)={\begin{cases}x&x\not =0\\1&x=0\end{cases}}$ Here we have what would be a continuous function (the function x) - we've just removed the one point where it would equal zero and moved that point somewhere else, in this case, to $y=1$ . You could come up with many other piecewise functions using these same ideas.