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

Jan 10, 2016

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

That is,

if $f ' ' \left(0\right) > 0$, the function is convex when $x = 0$
if $f ' ' \left(0\right) < 0$, the function is concave when $x = 0$

Here $f \left(x\right) = - {x}^{3} + 2 {x}^{2} - 4 x - 2$

Let $f ' \left(x\right)$ be the first derivative

$\implies f ' \left(x\right) = - 3 {x}^{2} + 4 x - 4$

Let $f ' ' \left(x\right)$ be the second derivative

$\implies f ' ' \left(x\right) = - 6 x + 4$

Put $x = 0$ in the second derivative i.e $f ' ' \left(x\right) = - 6 x + 4$.

$\implies f ' ' \left(0\right) = - 6 \cdot 0 + 4 = 0 + 4 = 4$
$\implies f ' ' \left(0\right) = 4$

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