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

MATH110 December 2010

• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q6 (c) • Q7 • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) • Q9 (e) • Q9 (f) • Q10 •

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

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[show] 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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[show] Solution
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 $(\infty ,1)\cup (1,\infty )$ .