Question

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

Answer 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`