MHB 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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The function f is defined as differentiable with f'(x) = x^2/(1+x^2) and f(0) = 0. The discussion focuses on verifying the inequality 0 ≤ f(x) ≤ x for all real numbers x. The solution to the corresponding ordinary differential equation y' = x^2/(1+x^2) with initial condition y(0) = 0 is y = x - arctan(x), which holds true for x ≥ 0. The analysis confirms that f(x) remains bounded by x for non-negative values, supporting the original inequality. The findings indicate that the function behaves consistently within the specified bounds across its domain.
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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.
 
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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