Open Question: Let F(x) = ʃ t-3 / t^2+7 dt (from 0 to x) for -oo <x<oo?

I assume you mean F(x) = ? (t-3) / (t^2+7) dt (t = 0 to x) ---> don't forget parentheses

F'(x) = (x-3) / (x^2+7)

F''(x) = (1*(x^2+7) - (x-3)(2x)) / (x^2+7)^2
F''(x) = (-x^2+6x+7) / (x^2+7)^2
F''(x) = -(x+1)(x-7) / (x^2+7)^2

a)

F has minimum value where F'(x) = 0 and F''(x) > 0

F'(x) = 0
(x-3) / (x^2+7) = 0
x = 3

F''(3) = -(3+1)(3-7) / (3^2+7)^2 = 16/16^2 = 1/16 > 0

Minimum value at x = 3

b)

F is increasing when F'(x) > 0
(x-3) / (x^2+7) > 0
x - 3 > 0 . . . . . . since (x^2+7) > 0 for all x
x > 3

F is decreasing when F'(x) < 0
x < 3

Increasing on interval (3, 8)
Decreasing on interval (-8, 3)

c)

F is concave up when F''(x) > 0
-(x+1)(x-7) / (x^2+7)^2 > 0
-(x+1)(x-7) > 0 . . . . . . since (x^2+7) > 0 for all x
-1 < x < 7

F is concave down when F''(x) < 0
x < -1 or x > 7

Concave up on interval (-1, 7)
Concave down on intervals (-8, -1) U (7, 8)

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