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 discussion focuses on the differentiable function f:R->R with the derivative f'(x) = [x^2 / (1 + x^2)] and the initial condition f(0) = 0. Participants analyze whether the inequality 0 ≤ f(x) ≤ x holds for all x in R. Additionally, they explore the equation y = c1 + c2.cos(y) with conditions c1 > 0 and 0 < c2 < 1, demonstrating that it has only one root not exceeding c1 + c2. The solution to the associated ordinary differential equation (ODE) is given as y = x - tan^(-1)(x), valid for x ≥ 0.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and inequalities
  • Familiarity with ordinary differential equations (ODEs) and their solutions
  • Knowledge of trigonometric functions and their properties
  • Basic understanding of limits and continuity in real analysis
NEXT STEPS
  • Study the properties of differentiable functions and their bounds
  • Learn about the existence and uniqueness theorems for ordinary differential equations
  • Explore the implications of trigonometric identities in solving equations
  • Investigate the behavior of the function y = x - tan^(-1)(x) for various x values
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Mathematicians, calculus students, and anyone interested in the applications of differentiation and ordinary differential equations in real analysis.

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.
 
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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