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

MATH110 December 2017

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[hide] 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.

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[show] Hint 1
Where is the function undefined?

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

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[show] Solution

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: $\color {blue}x\neq -1,0,{\frac {1}{2}}$