Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 36

g(x)=200+8x^3+x^4
differentiating,
g'(x)=24x^2+4x^3
g'(x)=4x^2(x+6)
Now let us find our critical numbers by setting g'(x)=0
4x^2(x+6)=0
solving above , x=0 and x=-6
Now let us check sign of g'(x) at test values in the intervals (-oo ,-6) , (-6,0) and (0,oo )
f'(-7)=4(-7)^2(-7+6)=-196
f'(-1)=4(-1)^2(-1+6)=20
f'(1)=4(1)^2(1+6)=28
Since f'(-7) is negative , function is decreasing in the interval (-oo ,-6)
f'(-1) is positive so function is increasing in the interval (-6,0)
f'(1) is positive so function is increasing in the interval (0,oo )
Since f'(-7) is negative and f'(-1) is positive , therefore Local minimum is at x=-6 . Local minima can be found by plugging in x=-6 in the function.
f(-6)=200+8(-6)^3+x^4=-232
Local minimum=-232 at x=-6
Now to find the intervals of concavity and inflection points,
g''(x)=4(x^2+(x+6)(2x))
g''(x)=4(x^2+2x^2+12x)
g''(x)=4(3x^2+12x)
g''(x)=12x(x+4)
set g''(x)=0
x=0 , x=-4
Now let us check the concavity in the intervals (-oo ,-4) , (-4,0) and (0,oo ) by plugging test values in g''(x).
g''(-5)=12(-5)(-5+4)=60
g''(-2)=12(-2)(-2+4)=-48
g''(1)=12(1)(1+4)=60
Since g''(-5) is positive , function is concave up in the interval (-oo ,-4)
g''(-2) is negative , so the function is concave down in the interval (-4,0)
g''(1) is positive , so the function is concave up in the interval (0,oo )
Since the concavity is changing so x=0 and x=-4 are the inflection points.