lighting quick question....

 

how do you show this? i dunno how to include abs values and stuff....

 

 

 

Thanks

 

Sarah

Well it's false so I wouldn't try to prove it, unless you're failing to mention what g is.

oops :S, my mistake, g is g(x) = (x+9)^(1/3) - x , and the interval is [2,3]

So, you know, by the MVT that there is y in the interval [u,v] such that

 

f(v)-f(u) = f'(y)(v-u) whenever v and u are in [2,3]

 

or |f(v)-f(u)| = |f'(y)||v-u|

 

now, what's the minimum value that f'(y) can be if y is some number in [2,3]?

when y = 3?

which is a minimum value of..... f'(3) = -0.9364

i mean f'(2) is the minimum = -0.9326

Sorry I meant to say what is the maximal value of |f'| on that interval for then

 

|f(v)-f(u)| = |f'(y)||v-u| < max{|f'|}|v-u|

 

so as long as the max of |f'| is less than a quarter we're ok.

oh ok, yes that makes sense.

 

thanks matt

 

-Sarah