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

I found \(f(0)=0\) but now I am not seeing how to proceed. Just a hint would be nice.

I further think I proved that the function is odd, though I think that doesn't really help me :/

use the definition of a derivative

the fact that it is odd does seem to help with the last part

\[f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}\]

limit f(x)-f(0) something...

okay so that is f'(0)=1, on to the last part
you say the oddness could help?

yes

can I use the derivative of an odd function is even?

don't need it

use the definition of a derivative again

that's what I'm doing, let me toy for a sec...

I found f(x) ;)

...if you would be so kind

if you know f'(x) and that f(0)=0 then finding f(x) is really easy :)

\[f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}\]

\[f(y)-f(x)=f(y)+f(-x)\]

does that help?

ok

yes

gotta go...fun problem...I'll post it later if you like

always a pleasure :)