f(x)=(2x+1)^3
a)find where f is increasing
b)find where f is decreasing
c)find the open intervals where f is concave up
d)find where f is concave down
e)x coordinates of the inflection points

Question

anonymous · Answer

whatever you do, DO NOT MULTIPLY THIS OUT! it looks just like 
$$f(x)=x^3$$ only shifted.  so it is always increasing.   you can use calculus if you like i suppose

anonymous · Answer

$$f'(x)=6(2x+1)^2$$ so only critical point is at 
$$x=-\frac{1}{2}$$ and since the derivative does not change sign there, it is neither a local max nor min.

anonymous · Answer

second derivative is not necessary since this one will clearly be concave down until the critical point and then concave up.  but if you need to write 
$$f''(x)=48x+24$$ and infection point is again at 
$$x=-\frac{1}{2}$$

turingtest · Answer

ha ha
"infection point"
sounds bad...