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

f(x)=(x^2-1)/(40x^3+x+1)
f'(x)=((40x^3+x+1)(2x)-(x^2-1)(120x^2+1))/(40x^3+x+1)^2
f'(x)=((80x^4+2x^2+2x)-(120x^4+x^2-120x^2-1))/(40x^3+x+1)^2
f'(x)=(80x^4-120x^4+2x^2-x^2+120x^2+2x+1)/(40x^3+x+1)^2
f'(x)=(-40x^4+121x^2+2x+1)/(40x^3+x+1)^2
differentiate again to get the second derivative,
f''(x)=((40x^3+x+1)^2(-160x^3+242x+2)-(2(40x^3+x+1)(120x^2+1))(-40x^4+121x^2+2x+1))/(40x^3+x+1)^4
f''(x)=(40x^3+x+1)((40x^3+x+1)(-160x^3+242x+2)-2(120x^2+1)(-40x^4+121x^2+2x+1))/(40x^3+x+1)^4
f''(x)=((-6400x^6+9680x^4+80x^3-160x^4+242x^2+2x-160x^3+242x+2)-(-9600x^6+29040x^4+480x^3+240x^2-80x^4+242x^2+4x+2))/(40x^3+x+1)^3
f''(x)=(3200x^6-19440x^4-560x^3-240x^2+240x)/(40x^3+x+1)^2
Refer the image and links, graphs of f(x) plotted in red color,f'(x) in blue color and f''(x) in green color.
From the graph,
f(x) is increasing in the interval about (-1.75,1.75)
f(x) is decreasing in the intervals (-oo ,-1.75) and (1.75,oo )
Local Maximum ~~ 0.01 at x ~~ 1.75
Local Minimum ~~ -0.01 at x ~~ -1.75
Inflection points are at x ~~ -2.5, x=0 , x~~ 0.2 , x ~~ 2.5
Function is concave up in the intervals about (-oo ,-2.5),(0,0.2)and (2.5,oo )
Function is concave down in the intervals about (-2.5,0) and (0.2,2.5)