Calculus: Early Transcendentals, Chapter 4, 4.6, Section 4.6, Problem 9

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)