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

1 Answer
Mar 7, 2017

The function is convex at $x = - 1$

Explanation:

We need

$\left(u v w\right) ' = u ' v w + v ' u w + w ' u v$

We must calculate the first and second derivatives

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

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

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

$= \left(x - 4\right) \left\{{x}^{2} - 2 x - 8 + {x}^{2} - 7 x + 12 + 2 {x}^{2} - 2 x - 12\right\}$

$= \left(x - 4\right) \left(4 {x}^{2} - 11 x - 8\right)$

$f ' ' \left(x\right) = \left(4 {x}^{2} - 11 x - 8\right) + \left(x - 4\right) \left(8 x - 11\right)$

$= 4 {x}^{2} - 11 x - 8 + 8 {x}^{2} - 43 x + 44$

$= 12 {x}^{2} - 54 x + 36$

Therefore,

$f ' ' \left(- 1\right) = 12 + 54 + 36 = 102$

As, $f ' ' \left(- 1\right) > 0$, we conclude that the function is convex