F(x) = 2sin(x)+x^2+5x+6. Why doesn't this function have any inflection points

To find inflection points, find double derivative and solve for f"(x)=0 or DNE:

f"(x) = -2sin(x) + 2
0 = -2sin(x) +2
1 = sin(x)
x = pi/2, 3pi/2 etc...

However, this function doesn't have any inflection points (you can graph it and see). How can I prove this mathematically? What am I doing wrong?

It is not sufficient that f ''(x) = 0 for there to be an inflection point. The concavity has to change. That's the issue with this function.

The second derivative is clearly zero many many (infinitely many) times. But, the sign of f '' never actually changes.

We know that -1 ≤ sin(x) ≤ 1 for all real x. Now note that

f ''(x) = 2 - 2sin(x) = 2(1 - sin(x)) ≥ 0

for all real x. The function is never concave down!

No change of concavity = no inflection points !!

x = pi/2 + 2 pi N

N = 1,2,3,.....