Dec 2, 2012 Thread starter #1 K kkafal New member Dec 2, 2012 2 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.
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.
Dec 2, 2012 Thread starter #2 K kkafal New member Dec 2, 2012 2 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.
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.
Dec 2, 2012 #3 chisigma Well-known member Feb 13, 2012 1,704 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. Click to expand... 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 $\chi$ $\sigma$
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. Click to expand... 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 $\chi$ $\sigma$