Science:Math Exam Resources/Courses/MATH110/December 2016/Question 07 (d)

MATH110 December 2016

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Question 07 (d)
Consider the function $f(x)={\frac {\sin(x)}{1-\cos(x)}}$. (d) Explain why there are no points on the curve $y=f(x)$ where the tangent line has slope $1$.

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
Recall that ${\textstyle -1\leq \cos(x)\leq 1}$ for any real number ${\textstyle x}$.

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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. Recall that the slope of the tangent line is the derivative of the function. By part (c), if ${\textstyle f'(x)=1}$, then $$-{\dfrac {1}{1-\cos(x)}}=1.$$ Rearranging, $$-1=1-\cos(x),$$ $$\cos(x)=2$$ However, ${\textstyle -1\leq \cos(x)\leq 1}$ for any ${\textstyle x}$. Thus, the above equation has no solution. In other words, there does not exist a value ${\textstyle x}$ so that ${\textstyle f'(x)=1}$. Answer: ${\textstyle {\color {blue}f'(x)=1{\mbox{ is equivalent to }}\cos(x)=2{\mbox{ which has no solution.}}}}$