Is #f(x) =(x+1)^3+3x^2+4x-3# concave or convex at #x=-1#?

The second derivative is positive, hence the curve is concave upwards at #x=-1#

Explanation:

Given -

#f(x)=(x+1)^3+3x^2+4x-3#

#f'(x)=3(x+1)^2(1)+6x+4#
#f'(x)=3(x+1)^2+6x+4#
#f''(x) = 6(x+1)+6#
#f''(x)=6x+6+6#
#f''(x)=6x+12#

At #f''(x) = 0; 6x+12=0#

#6x=-12#
#x=-12/6=-2#

At #x=-2# there is point of inflextion.

At #x=-1; f''(x)= 6(-1)+12=6>0#

At #x=-1# The second derivative is positive, hence the curve is concave upwards at #x=-1#