f(x)=((x+1)^3)-2
-Determine the number of turning points f has.
f'(x) = 3(x+1)^2
at turning points, f'(x) = 0
so
3(x+1)^2 = 0
x+1 = 0
x = -1
f(-1) = 0-2 = -2
since there was only one solution for x
there is only one turning point, namely (-1,-2)