Science:Math Exam Resources/Courses/MATH100 C/December 2024/Question 04 (b)

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MATH100 C December 2024

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5  •  Q6  •  Q7  •  Q8  •  Q9 (a)  •  Q9 (b)  •  Q10 (a)  •  Q10 (b)  •  Q10 (c)  •  Q11 (a)  •  Q11 (b)  •  Q11 (c)  •  Q12 (a)  •  Q12 (b)  •  Q12 (c)  •  Q12 (d)  •  Q13  •  Q14  •  

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  •  December 2023  •  December 2024  •  

Question 04 (b)
For this question, you do not need to justify your answers. Give the locations and types (jump, removable, or infinite) of all discontinuities of the function $f(x)=\frac{x+1}{x^2-1}$, or write “none” if there are none.

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
Similarly to part (a), determine the values of $x$ for which $x^2-1=0$ by factoring. Then examine whether any factors cancel with the numerator, and ask what type of discontinuity that indicates.

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
Following example in part (a), we can factor the denominator as follows $$\begin{array}{r}{x}^2-1=(x-1)(x+1).\end{array}$$ The function then becomes, $$\begin{array}{r}f(x)=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1},\end{array}$$ i.e., one of the factors corresponding to x=-1 cancels with the numerator. Thus, the function has a removable discontinuity at $x=-1$ and an infinite discontinuity at $x=1$.

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