# Produce graphs of f that reveal all the important aspects of the curve. Then use calculus to find the intervals of increase and decrease and the intervals of concavity.Produce graphs of f that...

Produce graphs of *f* that reveal all the important aspects of the curve. Then use calculus to find the intervals of increase and decrease and the intervals of concavity.

Produce graphs of *f* that reveal all the important aspects of the curve. Then use calculus to find the intervals of increase and decrease and the intervals of concavity. (Enter your answers in interval notation. Do not round your answers.)

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

a) Find the interval of increase.

b) Find the interval of decrease.

c) Find the interval where the function is concave up.

d) Find the interval where the function is concave down.

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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.