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
62since 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
62gallardo 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
11H'mmm.....actually I got confused bcoz, it asked for the maximum value of the function....
62There 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...