Question

Is `f(x)=-x^3+2x^2-4x-2` concave or convex at `x=0`?

Answer 1

If `f(x)` is a function, then to find that the function is concave or convex at a certain point we first find the second derivative of `f(x)` and then plug in the value of the point in that. If the result is less than zero then `f(x)` is concave and if the result is greater than zero then `f(x)` is convex.

That is,

if `f''(0)>0`, the function is convex when `x=0`
if `f''(0)<0`, the function is concave when `x=0`

Here `f(x)=-x^3+2x^2-4x-2`

Let `f'(x)` be the first derivative

`implies f'(x)=-3x^2+4x-4`

Let `f''(x)` be the second derivative

`implies f''(x)=-6x+4`

Put `x=0` in the second derivative i.e `f''(x)=-6x+4`.

`implies f''(0)=-6*0+4=0+4=4`
`implies f''(0)=4`

Since the result is greater then `0` therefore the function is convex.