anonymous
  • anonymous
Missing something. Functions. Let f : R -> R be a function such that : f( (x+y) /3) = [f(x) + f(y)] /3 then: Is f(x) differentiable in R? Is it continous? Is f(x)/x differentiable?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
f(x) = kx?
anonymous
  • anonymous
I know. It seems like that. But apparently not. As in the answer it's NOT differenciable in R.
anonymous
  • anonymous
Oh i missed some info. ** f(0) = 0 , f'(0) = 3

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
f(x)=3x
anonymous
  • anonymous
How is the function f(x)=kx not differentiable on/in R. Hmm \[\lim_{h \to 0} \frac{3(x+h) - 3x}{h}\]
anonymous
  • anonymous
Am I doing something wrong?
anonymous
  • anonymous
Did you differenciate-Put x=0---integrate? { To get the function} Or observe? :P-- That works here too . I got the same thing too. I dunno whats wrong.
anonymous
  • anonymous
@KingGeorge
anonymous
  • anonymous
Oh well. I guess the answer may be wrong then. All of us cant go wrong.
anonymous
  • anonymous
Thanks for Helping. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.