just a quick one...

[Sarahisme]

Hey can i ask if my answers to this question are correct...?

 

a) yes, because f'(x) = 0 at x = 0

b) yes, because there exists a tangent line at x = 0 (because of part a) and the concavity is of a different sign on both sides of x = 0 (i.e. f''(x) is -ve one left and +ve on right)

c) no it is not true, because f''(x) = 2 for all real x in the domain

d) no, as can been seen in this question

 

 

Sarah

[Dave]

Seems okay, apart from ©. I don't think your second derivative is quite correct.

[uncool]

f''(x) jumps from -2 to 2 at 0.

d is therefore anwered incorrectly. The answer is that the second derivative does necessarily vanish.

-Uncool-

[Sarahisme]

ok, so so part c should be

 

c) f''(x) is either -2 or 2, so not zero

and

d) no not necessarily

[matt grime]

Well, in c) you ought to say that the function only has a one sided second derivative at x=0, and some people, like me, would point out that therefore f'' doesn't exist at x=0.

[Sarahisme]

isnt the first derivative only one sided too? (at x = 0) ?? and therefore the first derivative does not exist at x = 0 ??

[Sarahisme]

lol, anyone? or am i way off track?

[Sarahisme]

nevermind, i am going crazy....

[Sarahisme]

i've got it now! problem solved