Science:Math Exam Resources/Courses/MATH110/April 2018/Question 02 (a)/Solution 1

Remember that the functions ${\textstyle 1-x}$ and ${\textstyle 1+\cos(x)}$ are each continuous everywhere.

First, we check the denominator is continuous on the real line. Note that a composite function of two continuous functions is also continuous on its domain. Remember that ${\sqrt {x}}$ is continuous on its domain $\{x:x\geq 0\}$. On the other hand, since ${\textstyle \cos(x)}$ is always at least ${\textstyle -1}$, ${\textstyle 1+\cos(x)}$ will always be at least zero. i.e., it is in the domain of the square root function. Therefore, the denominator (which is a composite of ${\sqrt {x}}$ and $1+\cos(x)$) is well-defined in the real line, so it is continuous on the real line.

Now, the only remaining thing to check is that the denominator is nonzero. ${\textstyle 1+\cos(x)}$ is nonzero if and only if ${\textstyle \cos(x)\neq -1}$, and ${\textstyle \cos(x)=-1}$ if and only if ${\textstyle x}$ is an odd integer multiple of ${\textstyle \pi }$. Since the numerator is nonzero for these values of ${\textstyle x}$, we conclude that ${\textstyle f(x)}$ is continuous for all ${\textstyle x}$ except ${\textstyle x=(2k+1)\pi }$, where ${\textstyle k}$ is an integer.

Answer: $f$ is continuous at any $\color {blue}x\neq (2k+1)\pi ,k\in \mathbb {Z}$.