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

1 Answer
Ori Kiko
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'(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]}

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