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

MATH110 December 2010

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Question 09 (d)
Let $f(x)={\frac {x+1}{x-1}}$. Find all critical points of $\displaystyle f$, if they exist.

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 the critical points of $f(x)$ are those x-values where the derivative $\displaystyle f'(x)$ is zero or undefined.

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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 start by calculating the derivative of $f(x)$, using the quotient rule: $f'(x)={\frac {(x-1)-(x+1)}{(x-1)^{2}}}={\frac {-2}{(x-1)^{2}}}$. The critical points of the function are the x-values where the derivative is either undefined or equal to zero. The derivative here is undefined when the denominator is equal to zero. Setting the denominator equal to zero... $(x-1)^{2}=0$ ...and solving for x, we get that $\displaystyle f'(x)$ is undefined when $x=1$. To find the remaining critical points, we set the derivative equal to zero and try solving for x. ${\frac {-2}{(x-1)^{2}}}=0$ As it turns out, there are no x-values that will satisfy this equation. Thus our only critical point is at $x=1$.

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