Science:Math Exam Resources/Courses/MATH110/December 2017/Question 05 (b)

MATH110 December 2017

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Question 05 (b)
(b) Consider the function $$g(x)={\begin{cases}\displaystyle {\frac {3x}{x^{2}-1}}&{\text{if }}x\leq 0\\&\\\displaystyle {\frac {x+3}{2x-1}}&{\text{if }}x>0.\end{cases}}$$ Find all points where ${\textstyle g}$ is continuous. If ${\textstyle g}$ is not continuous at a point, explain why by identifying which of the conditions listed above the function fails to satisfy.

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.

Hint 1
Where is the function undefined?

Hint 2
Consider also the point ${\textstyle x=0}$.

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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. Since a rational function is undefined at the point where the denominator is zero, we see that ${\textstyle g(x)}$ is not defined when $$x^{2}-1=0{\text{ and }}x\leq 0$$ or $$2x-1=0{\text{ and }}x>0,$$ that is, when ${\textstyle x=-1}$ and ${\textstyle x={\frac {1}{2}}}$. At these points ${\textstyle g(x)}$ is not defined hence is not continuous. On the other hand, ${\textstyle g(x)}$ changes formula across the point ${\textstyle x=0}$. Near this point, we have $$\lim \limits _{x\to 0^{-}}g(x)=\lim \limits _{x\to 0^{-}}{\dfrac {3x}{x^{2}-1}}={\dfrac {3\cdot 0}{0^{2}-1}}=0,$$ $$\lim \limits _{x\to 0^{+}}g(x)=\lim \limits _{x\to 0^{+}}{\dfrac {x+3}{2x-1}}={\dfrac {0+3}{2\cdot 0-1}}=-3,$$ $$g(0)={\dfrac {3\cdot 0}{0^{2}-1}}=0.$$ These three values exist, but one is not equal to the others. Therefore, ${\textstyle g(x)}$ is not continuous at ${\textstyle 0}$. So ${\textstyle g(x)}$ is continuous at all real numbers except ${\textstyle -1}$, ${\textstyle 0}$ and ${\textstyle {\frac {1}{2}}}$. To summarize. ${\textstyle g(x)}$ is continuous at all real numbers except ${\textstyle -1,0,{\frac {1}{2}}}$. ${\textstyle g(x)}$ is not continuous at $-1$ and ${\frac {1}{2}}$ because $g$ is not defined. ${\textstyle g(x)}$ is not continuous at 0 because the right hand limit is not equal to the left hand limit and the value $g(0).$ Answer: $\color {blue}x\neq -1,0,{\frac {1}{2}}$