# Produce graphs of f that reveal all the important aspects of the curve. 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 +(7/x^2)+(1/x^3) Find the interval of increase. Find the interval of decrease Find the interval where the function is concave up. Find the interval where the function is concave down.

An

This graph is betwee [-1,1]

`f'(x) = -1/x^2 - 14/x^3 - 3/x^4`

`f''(x) = 2/x^3 + 42/x^4 + 12/x^5`

The function is increasing when f'(x)>0.

f'(x) > 0 when `-1/x^2 - 14x^3 - 3/x^4 gt 0`

Multiply by `-x^4` to get

`x^2 + 14x + 3 lt 0`

We...

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An

This graph is betwee [-1,1]

`f'(x) = -1/x^2 - 14/x^3 - 3/x^4`

`f''(x) = 2/x^3 + 42/x^4 + 12/x^5`

The function is increasing when f'(x)>0.

f'(x) > 0 when `-1/x^2 - 14x^3 - 3/x^4 gt 0`

Multiply by `-x^4` to get

`x^2 + 14x + 3 lt 0`

We get `(x-(-14+-sqrt(14^2 - 4(1)(3)))/2)` .

This happens when `x = (x+7+sqrt(184)/2)` or `(x+7-sqrt(184)/2)`

Simplifying we get `(x+7+sqrt(46))(x+7-sqrt(46))` .  This is <0 (and the function is deceasing) on the intervals `(-oo,-7-sqrt(46)), (-7+sqrt(46, 0), (0, +oo)` .

Increasing >0 on the intervals `(-7-sqrt(46),-7+sqrt(46))`

Concave up when f''(0) > 0 and concave down when f''(0) < 0.

`f''(x) = 2/x^3 + 42/x^4 + 12/x^5 gt 0`

`2x^2 + 42x + 12 gt 0`

`x^2 + 21x + 6 gt 0`

`x = (-21+-sqrt(21^2-4(1)(6)))/2 = (-21+-sqrt(417))/2`

`(x-(-21-sqrt(417))/2)(x-(-21+sqrt(417))/2)lt0`

When `(-oo,(-21-sqrt(417))/2)` , `((-21+sqrt(417))/2,0)` the function is concave down,

When `((-21-sqrt(417))/2,(-21+sqrt(417))/2)` and `(0,oo)` it is concave up.

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