Q)Show that the function f colon R rightwards arrow left curly bracket x element of space R space colon space minus 1 less than x less than 1 right curly bracket space space d e f i n e d space b y space f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus open vertical bar x close vertical bar end fraction comma space x element of R space i s space o n t o.

Asked by araima2001 | 4th Jun, 2017, 07:49: PM

Expert Answer:

begin mathsize 16px style straight f left parenthesis straight x right parenthesis equals fraction numerator straight x over denominator 1 plus open vertical bar straight x close vertical bar end fraction
Case space 1 colon
minus 1 less than straight x less than 0
rightwards double arrow straight f left parenthesis straight x right parenthesis equals straight y equals fraction numerator straight x over denominator 1 minus straight x end fraction
rightwards double arrow straight y minus yx equals straight x
rightwards double arrow straight y equals xy plus straight x
rightwards double arrow straight y equals straight x open parentheses 1 plus straight y close parentheses
rightwards double arrow straight x equals fraction numerator straight y over denominator 1 plus straight y end fraction
Similarly comma space for space 0 less than straight x less than 1
straight x equals fraction numerator straight y over denominator 1 minus straight y end fraction
rightwards double arrow straight x equals fraction numerator straight y over denominator 1 plus open vertical bar straight y close vertical bar end fraction
HEnce comma space function space is space onto. end style

Answered by Sneha shidid | 5th Jun, 2017, 12:38: PM