if f(x) is concave up on (a,b), then f''(x) > 0 on (a,b)
f''(x) > 0 on (0,5) ...
The graph of f(x) = [0,x] integral (15t^2 - 2t^3 + 24) dt is concave up on (a,b). Find b - a
so would f(x) = 15x^2 - 1/2x^4 + 24x?
and then I take the derivative of that which would be 15x^2 - 2x^3 + 24 and set it equal to zero? this is confusing . . .