H'mmm.....actually I got confused bcoz, it asked for the maximum value of the function....
Suppose f is a real valued differentiable function defined on [1,\propto) with f(1)=1.
Given the function satisfies the relation f'(x)=\frac{1}{x^2+f^2(x)} - so now find the maximum value of f(x) for x\ge 1.
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6 Answers
since f(1)=1
and f(x)>0 for all x>1,
f'(x)<\frac{1}{x^2+1}
Now integrate from 1 to x, to get
f(x)-f(1) < tan-1x-tan-11
but tan-1x is again less than pi/2
so f(x)-f(1) < pi/2-pi/4
so f(X)<1+pi/4
gallardo see a simple function 1/x in the interval (1,infinity)
its derivative is -ve always..
But it has a minimum value going to zero
There is a small thing...
The quesiton is not well written..
In the sense that the right question is not to find the maxima but to prove that it is bounded above by that number...