anonymous
  • anonymous
Given the function... is it derivable at the point P(0,-1) ?
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
helder_edwin
  • helder_edwin
use the definition: the function is derivable at x=0 if \[ \large \lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)+1}{x} \]
helder_edwin
  • helder_edwin
this limit has to exist. so u now turn to one-sided limits.

Looking for something else?

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

More answers

anonymous
  • anonymous
Oh yeeah, thank you so much. After you use the rules you forget the definition.
helder_edwin
  • helder_edwin
u r welcome
anonymous
  • anonymous
actually it is easier than that, although of course @helder_edwin is correct
anonymous
  • anonymous
just take the derivative of each piece, and replace \(x\) by \(0\) if you get the same answer, then yes, if you get a different answer, then no
anonymous
  • anonymous
you can pretty much do it with your eyeballs first one is 3 second one is \(10x+3\)and when you replace \(x\) by 0 in both you get 3
anonymous
  • anonymous
haha thanks :) Not everyone can do it with their eyeballs ;)

Looking for something else?

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