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

MATH110 April 2018

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Question 02 (a)
Determine for what values of ${\textstyle x}$ is ${\textstyle f(x)={\frac {1-x}{\sqrt {1+\cos(x)}}}}$ continuous.

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
Remember that when $f(x)$ and $g(x)$ are continuous at $x=a$ , so is ${\frac {f(x)}{g(x)}}$ if $g(a)\neq 0$ .

Hint 2
Check that the denominator is nonzero and that the value inside the square root is nonnegative.

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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. 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 $\color {blue}x\neq (2k+1)\pi ,k\in \mathbb {Z}$ .