# Findregions of increase/decrease, stationary points, directions of concavity, points of inflection and asymptotes of the functionF(x)=x+1/x-1,x not 0

### 2 Answers | Add Yours

Sometimes a question is mislabelled, and occasionlly brackets are missed. In this case, suppose the function was actually supposed to be `f(x)={x+1}/{x-1}`

The the vertical asymptote is at x=1 and the horizontal asymptote is at y=1.

There is an x-intercept at x=-1 and a y-intercept at y=-1.

The first derivative becomes

`f'(x)={(x-1)-(x+1)}/{(x-1)^2}`

`={-2}/{(x-1)^2}`

and the second derivative is

`f''(x)=4/{(x-1)^3}`

This is is concave up for x>1 and concave down for x<1. The function is also decreasing for all x except x=1 since that is a VA. It also means there are no stationary points, since the first derivative never vanishes.

The function is entirely monomials, so finding derivatives is done using the power rule.

`f(x)=x+x^{-1}-1`

`={x^2+1-x}/x`

`f'(x)=1-x^{-2}`

`={x^2-1}/x^2`

`={(x-1)(x+1)}/x^2`

`f''(x)=3/x^3`

The stationary points are where the first derivative vanishes, which is at x=1, and x=-1.

For the interval x>1, `f'(x)>0` so this region is increasing. For 0<x<1, `f'(x)<0` so this region is decreasing. For -1<x<0, `f'(x)>0` so this region is increasing. Finally, for x<0, `f'(x)<0` so this region is decreasing.

The points of inflection are where the second derivative vanishes. This never happens, although `f''(x)>0` for x>0 and `f''(x)<0` for x<0 so the interval x>0 is concave up and the interval x<0 is concave down.

From the original function, we see that there is an oblique asymptote at y=x-1. Also, there is a vertical asymptote at x=0 since the denominator is zero there.