# increasing and decreasing functions

• Nov 5th 2009, 12:12 AM
bosmith
increasing and decreasing functions
I am to determine the interval on which f(x)=x-x^-2 is increasing. I came up with the derivative of f(x) =x-x^-2 as f(x)=1+2/x^3. Having said that, the graph on the left side of the y axis seems to travel in a positive direction until it gets to the y axis, then goes to -infinity. The graph on the right side of the y axis seems to go from o to infinity. Am I doing this correctly? Is the increasing interval (o, to infinity)?
• Nov 5th 2009, 03:01 AM
HallsofIvy
Quote:

Originally Posted by bosmith
I am to determine the interval on which f(x)=x-x^-2is increasing. I came up with the derivative of f(x) =x-x^-2

Poorly written! You do not mean to say that f'(x) is $x-x^{-2}$ and that is what "=" means! You mean $f(x)= x-x^{-2}$

Quote:

as f(x)=1+2/x^3. Having said that, the graph on the left side of the y axis seems to travel in a positive direction until it gets to the y axis, then goes to -infinity. The graph on the right side of the y axis seems to go from o to infinity. Am I doing this correctly? Is the increasing interval (o, to infinity)?
Well, you've already done the work. The function is increasing as long as $f'(x)= x- x^{-2}> 0$. Since $x^2> 0$ for all real (non-zero) x, we can multiply on both sides by $x^2$ and get [tex]x^3- x> 0[\math]. That factors as $x(x^2- 1)= x(x-1)(x+1)> 0$. The simplest way to determine where the derivative is positive or negative is to recognize that it can change from one to the other only where it is negative. And that occurs only at x=-1, x= 0, and x= 1. Choose one value for x in each of the intervals $-\infty< x< -1$, -1< x< 0, 0< x< 1, and $1< x< \infty$ to see whether the derivative is positive or negative in that interval.