ths toh i know
but how to prove that f'(X)<0 for proving it to be dec func
how to prove that (xcosx-sinx)/x2 is <0
Ans) Let f(x) = sinxx
Since f(x) is dec in the interval (0 , ∩/2)
Therefore, f(0) > f(x) > f (∩/2)
That implies, 1 > sinxx > 1∩/2 = 2 / ∩
ths toh i know
but how to prove that f'(X)<0 for proving it to be dec func
how to prove that (xcosx-sinx)/x2 is <0
x2>0 for x>0
so only thing left to prove is xcosx-sinx<0
which is true iff sinx>xcosx
=> tanx>x
which is true ...(See graph)
one more
prove taht f(x)=(1+1x)1/x is monotonically increasing in its domain
and draw its graph also :)