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

MATH110 December 2010

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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!

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

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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. 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.

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