Hey! We 've verified this expert answer for you, click below to unlock the details :)
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.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
Okay. Lemme walk u through it..
If we have the graph of a function, it is continuous if it is differentiable at all points i.e. it has no breaks, sharp corners, etc.
if lim of f(x) as x approaches a from the left = the limit as x approaches a from the right, the limit exists.
if the limit exists at a, then the function is continuous at a.
[square brackets] mean that the number is included in the interval
(round brackets) mean that the point is not included in the interval.
Carry on @Euler271
@allank so when u say sharp corners do you also mean by turning points where it slowly rises and then falls
No, the function is differentiable there. |dw:1369549739726:dw|
@allank so from this graph how do we determine where the graph is continous??? Sorry for hassling u again and again
@Euler271 do u think you could help me solve this question as well
No prob. So avoid intervals where we breaks like: |dw:1369550343320:dw|
And vertical asymptotes like: |dw:1369550390687:dw|