On the interval [1.3,2],
what is the min of f'(x) and max of f''(x)?
f'(x) = 1/(x-1) -sin (x-1)
f"(x) = -1/(x-1)^2 - cos(x-1)

Question

On the interval [1.3,2],
what is the min of f'(x) and max of f''(x)?
f'(x) = 1/(x-1) -sin (x-1)   
f"(x) = -1/(x-1)^2  - cos(x-1)

zzr0ck3r · Answer

the derivative of f' is f'' so set f'' = 0 and solve for x to answer the first one, the find f''' and set f'''=0 for the second part.

anonymous · Answer

i thought that would give the max for both.
is finding the min the same?

zzr0ck3r · Answer

its the same process, you are finding these critical points, but you should check to see if they are mins or maxs but seeing how the problem told you what they will be, you may or may not need to do that.

anonymous · Answer

nvm just remembered what the derivate set to 0 means

anonymous · Answer

alright yea thanks!

zzr0ck3r · Answer

you can plug in numbers next to the x to see if its a min or max or look at the functions second derivative (this would mean f'''' for part 2) to see concavity.