Science:Math Exam Resources/Courses/MATH110/December 2010/Question 09 (e)

MATH110 December 2010

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Question 09 (e)
Let $f(x)={\frac {x+1}{x-1}}$. Determine where $\displaystyle f$ is increasing and where it is decreasing.

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
A function is increasing on an interval if its derivative is positive on that interval, and decreasing if its derivative is negative.

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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. We know that where a function is increasing or decreasing depends on the sign of its derivative. Therefore we consider the derivative we found in part (d). $f'(x)={\frac {-2}{(x-1)^{2}}}$ Note that the sign of the numerator is always negative because the numerator is a constant. Thus the only place the sign of the derivative can change is in the denominator. However, because the x-value in the denominator is squared, the denominator will always be positive. Since the numerator is always negative and the denominator always positive, the entire derivative will be negative for all x for which it is defined. This means that the function $f(x)$ is decreasing, on the interval $(\infty ,1)\cup (1,\infty )$.

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