Applications of differentiation

[abby_07]

i need help finding all relative extreme and points of inflection
this is as far as i can go ( well at least i think i am doing it correctley)
f(x)=x(x+1)^1/2
f'(x)=(x+1)^1/2+x/2(x+1)^1/2=0
critical #= -2/3
f"(x)= 1/2(x+1)^1/2 +[(x+2)/4(x+1)^3/2]
f"(-2/3)= a positive number therefore it is a relative min.
relative min. is (-2/3,-2(3)^1/2 /9)

I dont know if those are all my critical numbers and i do not know how to find the inflection points
can you help please

 

[galactus]

To find inflection points, set f''(x)=0 and solve for x.

f''(x) has to change direction there, though.

f(x) is discontinuous at x=-4/3, which is f''(-4/3)=0