# `f(x) = 1 + 1/x + 8/(x^2) + 1/(x^3)` Produce graphs of `f` that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and inter- vals of concavity, and...

`f(x) = 1 + 1/x + 8/(x^2) + 1/(x^3)` Produce graphs of `f` that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and inter- vals of concavity, and use calculus to find these intervals exactly.

*print*Print*list*Cite

### 1 Answer

See the attached graph and links. Graph is drawn in several ranges to have clarity.

**From the graph**,

f is decreasing in the intervals about (- `oo` , -16),(-0.2 ,0) and (0,`oo` )

f is increasing in the interval about (-16,-0.2)

Local minimum value f(-16) `~~` 0.97

Local maximum value f(-0.2) `~~` 72

f is concave down on (-`oo` ,-24) and (-0.25,0)

f is concave up on (-24,-0.25) and (0,`oo` )

Let's differentiate to find exact intervals,

Inflection point at (-24,0.97) and (-0.25 ,-60)

`f'(x)=(-1/x^2)+8(-2)x^-3+(-3)x^-4`

`f'(x)-1/x^2-16/x^3-3/x^4`

`f'(x)=-1/x^4(x^2+16x+3)`

Now to find critical points set f'(x)=0

`-1/x^4(x^2+16x+3)=0`

`x^2+16x+3=0`

solving by quadratic formula,

`x=(-16+-sqrt(16^2-4*3))/2=(-16+-sqrt(244))/2`

`x=-8+-sqrt(61)`

`x=-15.81 , x=-0.19`

Now let's find the sign of f' in the intervals (-`oo` ,-15.81) , (-15.81,-0.19) ,(-0.19,0) and (0,`oo` ) by plugging in test points.

`f'(-16)=-0.000045`

`f'(-1)=12`

`f'(-0.1)=-14100`

`f'(1)=-20`

Since f'(-16), f'(-0.1) and f'(1) are negative , so f is decreasing in the intervals (-`oo` ,-15.81) , (-0.19,0) and (0,`oo` )

Since f'(-1) is positive , so f is increasing in the interval (-15.81,-0.19)

Now to find the concavity of f let's find the second derivative,

`f''(x)=-((x^4(2x+16)-(x^2+16x+3)(4x^3))/x^8)`

`f''(x)=(-x^3/x^8)(2x^2+16x-4x^2-64x-12)`

`f''(x)=-1/x^5(-2x^2-48x-12)`

`f''(x)=2/x^5(x^2+24x+6)`

Now let's find the intervals of concavity by setting f''(x)=0

`2/x^5(x^2+24x+6)=0`

`x^2+24x+6=0`

solve using quadratic formula,

`x=(-24+-sqrt(24^2-4*6))/2=(-24+-sqrt(552))/2`

`x=-12+-sqrt(138)`

`x=-23.75 , x=-0.252`

Now let's check the sign of f'' in the intervals (-`oo` ,-23.75) , (-23.75,-0.25) , (-0.25,0) and (0,`oo` )

`f''(-24)=2/(-24)^5((-24)^2+24(-24)+6)=-0.000001`

`f''(-1)=2/(-1)^5((-1)^2+24(-1)+6)=34`

`f''(-0.1)=2/(-0.1)^5((-0.1)^2+24(-0.1)+6)=-722000`

`f''(1)=2/1^5(1^2+24(1)+6)=62`

Since f''(-1) and f''(1) are positive, so the function is concave up in the intervals (-23.75,-0.25) and (0,`oo` )

Since f''(-24) and f''(-0.1) are negative , so the function is concave down in the intervals (-`oo` ,-23.75) and (-0.25,0)

**Sources:**