Concavity

[Becky4paws]

Could you double check me on this...I can't figure out where concavity is downward.

Determine where the given function in increasing/decreasing, concave up/concave down, relative extrema and inflection points.

f(x) = (x-2)^4

f'(x) = 4(x-2)^3
f"(x) = 12(x-2)^2

critical points (where f'(x) = 0)
f'(x) = 4(x-2)^3
0 = 4(x-2)^3
0 = (x-2)
2 = x

f(2) = 0
critical point (2,0)

Inflection Point (where f"(x) = 0 and concavity changes)
f"(x) = 12(x-2)^2
0 = 12(x-2)^2
0 = (x-2)
2=x

f(2) = 0
Inflection Point (2,0)

According to my points chart:

If x is 1: f(1) = 1, f'(1) = -4, f"(1) = 12
If x is 2: f(2) = 0, f'(2) = 0, f"(2) = 0
If x is 3: f(3) = 1, f'(3) = 4, f"(3) = 12

Increasing @ x>2
Decreasing @ x<2
Maximum: none
Minimum: (2,0)
Concave Up: all x except 2?
Concave Down: none?

Thanks for the help.

 

[tkhunny]

Where's your question? It looks to me like you have it.

 

[Becky4paws]

So there would NOT be any concave downward?

Concave Up @ all x except 2 and no concave down?

 

[trojan]

Yes Becky there is no concave down at all on this fucntion as you have shown with your work. The other way to easily confirm this is simply graph f(x) = (x-2)^4. If you graph this function you can see that it is an upward parabola, which is concave up. Never on the graph of f(x) do you see any downward concaves.