Hmm...That's really hard work... :)
.. and yes.. hopefully.:)
Let g'(x)>0 and f'(x)<0 for all x ε R
then prove that g(f(x+1))<g(f(x-1))
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7 Answers
kaymant
·2009-08-28 10:12:26
g'(x)>0 => g(x) is strictly increasing
f'(x)<0 => f(x) is strictly decreasing
Also for all real x,
x+1 > x-1
so how do f(x+1) and f(x-1) compare?
Lokesh Verma
·2009-08-28 10:17:23
yup i come online in the day time...
btw i will try to give you a call and visit you tomorrow... (Hopefully finally :)
Unfortunately the classes continute till late and having worked for 16 odd hours (with the travel) we feel very tired :)