Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 37

h(x)=(x+1)^5-5x-2
differentiating,
h'(x)=5(x+1)^4-5
Now let us find the critical numbers,
5(x+1)^4-5=0
5x(x+2)(x^2+2x+2)=0
solving above ,
x=0 , x=-2 , x=-1+i , x=-1-i
ignore the points that are complex
Now let us check the sign of f'(x) in the intervals (-oo ,-2) , (-2,0) and (0,oo )
Now let us take a test point in each interval
f'(-3)=5(-3+1)^4-5=75
f'(-1)=5(-1+1)^4-5=-5
f'(1)=5(1+1)^4-5=75
so f'(-3) is positive therefore function is increasing in the interval(-oo ,-2),
f'(-1) is negative therefore function is decreasing in the interval (-2,0)
f'(1) is positive therefore function is increasing in the interval (0,oo )
Now to find the local extrema plug in the critical numbers in the function
f(-2)=(-2+1)^5-5(-2)-2=7
f(0)=(0+1)^5-5(0)-2=-1
so Local Maximum=7 at x=-2
Local Minimum=-1 at x=-1
Now to find the intervals of concavity and inflection points let us find the second derivative of the function.
f''(x)=20(x+1)^3
Now let us find the point of inflection,
20(x+1)^3=0
solving above, x+1=0 , x=-1
Now let us check the concavity of the function by taking test point in the intervals (-oo ,-1) and (-1,oo )
f''(-2)=20(-2+1)^3=-20
f''(1)=20(1+1)^3=160
f''(-2) < 0 , so function is concave downward in the interval (-oo ,-1)
f''(1) > 0 , so function is concave upward in the interval (-1,oo )
So the inflection point is at x=-1 as the concavity changes.