Given `f(x)=1+1/x+9/x^2+1/x^3` :

First rewrite as `f(x)=1+x^(-1)+9x^(-2)+x^(-3)`

(1) To find the intervals of increase and decrease we find where the first derivative is positive or negative. The intervals we seek are between the points where `f'(x)=0` or `f'(x)` fails to exist.

`f'(x)=-1/x^2-18/x^3-3/x^4=-(x^2+18x+3)/x^4`

`f'(x)=0==>x^2+18x+3=0`

`==>x=-9+-sqrt(78)` or `x~~-.1682,x~~-17.8317`

Also, the derivative...

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Given `f(x)=1+1/x+9/x^2+1/x^3` :

First rewrite as `f(x)=1+x^(-1)+9x^(-2)+x^(-3)`

(1) To find the intervals of increase and decrease we find where the first derivative is positive or negative. The intervals we seek are between the points where `f'(x)=0` or `f'(x)` fails to exist.

`f'(x)=-1/x^2-18/x^3-3/x^4=-(x^2+18x+3)/x^4`

`f'(x)=0==>x^2+18x+3=0`

`==>x=-9+-sqrt(78)` or `x~~-.1682,x~~-17.8317`

Also, the derivative fails to exist at x=0. So we check the intervals:

`(-oo,-9-sqrt(78)):` try a test value such as x=-18. `f'(-18)~~-2.8578x10^(-5)<0` so the function is decreasing on this interval.

`(-9-sqrt(78),-9+sqrt(78)):` try a test value such as x=-1. `f'(-1)=14>0` so the function is increasing on this interval.

** The point at `x=-9-sqrt(78)` is a local minimum. The point at `x=-9+sqrt(78)` is a local maximum**

`(-9+sqrt(78),0):` try a test value such as x=-.1. `f'(-.1)=-12100<0` so the function is decreasing on this interval.

`(0,oo):` It is clear that the first derivative is negative for all x>0, so the function decreases on this interval.

(2) To find the intervals of concavity we examine the second derivative:

`f''(x)=2/x^3+54/x^4+12/x^5=(2(x^2+27x+6))/x^5`

The second derivative is zero at `x=-27/2+-sqrt(705)/2` . The second derivative fails at x=0. So we try the intervals:

`(-oo,-27/2-sqrt(705)/2):` try a test value such as x=-27: `f''(-27)~~-8.363x10^(-7)<0` so the function is concave down on this interval.

`(-27/2-sqrt(705)/2,-27/2+sqrt(705)/2)` `f''(-2)=2.75>0` so the function is concave up on this interval.

`(-27/2+sqrt(705)/2,0):` `f''(-.1)=-662000<0` so the function is concave down on this interval.

`(0,oo)` `f''(1)=68>0` so the function is concave up on this interval.

** `x=-27/2+-sqrt(705)/2` are inflection points **

Some graphs to show the interesting features:

The grapher in this program is unable to show the details around x=-17.83. There is a local minimum there.

The function is concave down, up, down; vertical asymptote; then concave up.

The function decreases, increases, decreases; vertical asymptote; then decreases.