Question

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

Answer 1

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'(x)=(-4xsqrt(-x^2+15x-36)-2x+15)/(2sqrt((x-12)(3-x)))`

and now let's derivate again,
`f''(x)=-2-(-2x+15)^(2)/(4(x-12)(3-x)sqrt((x-12)(3-x)))-1/sqrt((x-12)(3-x))`

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]}