Derivative of f:R->R with f'(x)=[x^2/1+(x^2))], f(0)=0 and its Bounds on R

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In summary, the conversation discusses a question on applications of differentiation and asks to show that the given equation has only one root. The conversation also mentions an ODE with separable variables and its solution. It is noted that the solution only satisfies the given relation for $x \ge 0$.
  • #1
kkafal
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f:R->R is differentiable and f'(x)=[x^2/1+(x^2))] and f(0)=0.
check whether 0<=f(x)<=x for all x belonging R.
Thanks.
 
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A question on applications of differentiation

Show that the equation y=c1+c2.cos(y) such that c1>0 and 0<c2<1 , has only one root not more than c1+c2.
 
  • #3
kkafal said:
f:R->R is differentiable and f'(x)=[x^2/1+(x^2))] and f(0)=0.
check whether 0<=f(x)<=x for all x belonging R.
Thanks.

In term of ODE we can write [if I undestood correctly...] ...

$\displaystyle y^{\ '}= \frac{x^{2}}{1 + x^{2}}\ ,\ y(0)=0$ (1)

The (1) is an ODE with separable variables and has as solution...

$\displaystyle y= x - \tan^{-1} x$ (2)

... so that Yor relation is satisfied... but only for $x \ge 0$... Kind regards$\chi$ $\sigma$
 

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