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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

the interval is [0,4]

Ok, you tell us: is \(|x-1|\) continuous on \([a,b]\) and differentiable on \((a,b)\)?

Hint:
|dw:1358042258899:dw|
and try to answer questions by oldrin.

i got to that part but does the vertical asymptote have anything to do with it?

The main questions are:
is f(x) continuous on (0,4), and
is f(x) differentiable on [0,4]

yes to both

no because it is 0

If it is zero, then the answer is yes, because zero is a number.

it is 1-1/ |1-1|

Is the derivative really zero?

\[f'(x)=\frac{ x-1 }{ |x-1| }\]

For the derivative to exist, the left and right limits of f(x) must be equal.

...exist and equal.

Would you propose the left limit, i.e. f'(1-), and the right limt f'(1+).

Yes that is correct.

they are all undefined?