Science:Math Exam Resources/Courses/MATH100/December 2012/Question 04 (a)

MATH100 December 2012

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Question 04 (a)

Short-Answer Questions. Questions 1-4 are short-answer questions. Put your answers in the boxes provided. Simplify your answers as much as possible, and show your work. Each question is worth 3 marks, but not all questions are of equal difficulty. Let a be a constant, and define $f(x)={\begin{cases}x^{2}+a&{\text{if }}x\leq 1\\2x-3&{\text{if }}x>1\end{cases}}$ Find the value of a for which f is continuous everywhere.

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
Since polynomials themselves are continuous everywhere, the only place where the function can be discontinuous is at the break point $x=1$. So we need to choose a so that the limit at 1 exists, or in other words that the left and right hand limits at 1 are equal.

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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. To make this function continuous everywhere, we need $\lim _{x\to 1}f(x)=f(1)=(1)^{2}+a=1+a$ In particular, we require that the limit on the left exists. To do this, check left and right hand limits and find a value of a that makes them equal. First $\lim _{x\to 1^{+}}f(x)=\lim _{x\to 1^{+}}x^{2}+a=1^{2}+a=1+a$ and for the other side $\lim _{x\to 1^{-}}f(x)=2x-3=2(1)-3=-1$ Thus we need 1 + a = -1 and so a = -2.

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