mean value thorem help please

Question

anonymous · Answer

just type it into google.

anonymous · Answer

show that the equation x^3-15x+c=0 has at most one root in the interval {2,2]

anonymous · Answer

$$
\exists c\;f(b)-f(a) = f'(c)(b-a)
$$

anonymous · Answer

the difinition is not my question here

anonymous · Answer

the book says
f(r)'=3r^2-15. since r is in (a,b) which is contained in [-2,2], we have |r|<2 so r^2<4. it follows that 3r^2-15< 3*4-15=^3<0

anonymous · Answer

$(2,2]$?  That's an empty interval.

anonymous · Answer

why its r^2=4??

anonymous · Answer

sorry its [-2,2]

anonymous · Answer

$$|r|<2\implies |r|^2=|r|\times |r|<2\times |r| < 2\times 2=4
$$

anonymous · Answer

o... thank you

anonymous · Answer

I still have no idea how their proof works though.

anonymous · Answer

what is your proof way to show work?

anonymous · Answer

Well I would straight out find the roots.  I wouldn't think to use mean value theorem.

anonymous · Answer

Do you understand their proof?

anonymous · Answer

mmm.. not 100% . i think i need more practice!

anonymous · Answer

It seems they're trying to show that the derivative is always negative on the interval.  If it is, then you can only pass through the $x$ axis once.

anonymous · Answer

See the derivative has change if you want to turn around and go back upward.

anonymous · Answer

Mmm .. yes. are you also familiar with increasing and decreasing intervals questions?

anonymous · Answer

Sure

anonymous · Answer

I wiill post the question 
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