anonymous
  • anonymous
Need help with calculus question quickly!
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
zepdrix
  • zepdrix
\[\large\rm |x-3|=\cases{(x-3),\qquad x\ge3\\ \rm -(x-3),\quad x\lt3}\]
zepdrix
  • zepdrix
So when we're on the `left side of 3`, Our |x-3| is -(x-3), giving us a function of:\[\large\rm t(x)=2^x-\frac{-(x-3)}{(x-3)}\]Err I guess it would be better to do this with limits. Approach 3 from the left, and seperately approach 3 from the right and see if the two pieces connect. If they don't, we have a jump discontinuity.

Looking for something else?

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

More answers

zepdrix
  • zepdrix
\[\large\rm \lim_{x\to3^-}t(x)=2^{x}-\frac{-(x-3)}{(x-3)}\]Do a cancellation before plugging the 3 in :)
anonymous
  • anonymous
Thanks! @zepdrix
zepdrix
  • zepdrix
Did you figure out out Princess Jasmine? :o
anonymous
  • anonymous
Yes. I said it was a jump discontinuity. I just needed a starting point.
anonymous
  • anonymous
@zepdrix

Looking for something else?

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