# Is f(x) =sqrt((x-12)(-x+3))-x^2 concave or convex at x=6?

Nov 16, 2017

Convex.

#### Explanation:

To know if a function is concave or convex you use the second derivative and you substitute $x$ by the point you want to know if it's concave or convex, in this case $x = 6$, if the number we get is positive it means that the function in that point is concave, if its negative it means that it's convex, and if it's $0$ it means that point is an inflection point.

Let's derivate one time,
$f ' \left(x\right) = \frac{- 4 x \sqrt{- {x}^{2} + 15 x - 36} - 2 x + 15}{2 \sqrt{\left(x - 12\right) \left(3 - x\right)}}$

and now let's derivate again,
$f ' ' \left(x\right) = - 2 - {\left(- 2 x + 15\right)}^{2} / \left(4 \left(x - 12\right) \left(3 - x\right) \sqrt{\left(x - 12\right) \left(3 - x\right)}\right) - \frac{1}{\sqrt{\left(x - 12\right) \left(3 - x\right)}}$

it only remains substituting,
f''(6)≈-2.265

and so the number we get is negative, it means that the funcion is convex, as we can see in the graph below:
graph{sqrt((x-12)(3-x))-x^2 [-3.25, 42.34, -21.04, 1.77]}