f(x)={x}/lxl=[x]-x/lxl find the domain  and range

Asked by Kalyanrao Chavan | 30th May, 2014, 04:41: PM

Expert Answer:

C o n s i d e r space t h e space g i v e n space f u n c t i o n space f open parentheses x close parentheses equals open square brackets x close square brackets minus fraction numerator x over denominator open vertical bar x close vertical bar end fraction ; W e space n e e d space t o space f i n d space t h e space d o m a i n space a n d space r a n g e space o f space f left parenthesis x right parenthesis. D o m a i n space o f space f space i s space t h e space s e t space o f space a l l space v a l u e s space o f space x space t h a t space x space c a n space t a k e. S i n c e space d i v i s i o n space b y space z e r o space i s space n o t space m e a n i n g f u l comma space x space c a n space a s s u m e space a l l space v a l u e s space e x e c e p t space x equals 0 ; space t h u s space x element of R minus open curly brackets 0 close curly brackets. T h e r e f o r e space d o m a i n space o f space f space i s space R minus open curly brackets 0 close curly brackets.  R a n g e space o f space f space i s space t h e space s e t space o f space a l l space v a l u e s space o f space f open parentheses x close parentheses L e t space y equals open square brackets x close square brackets minus fraction numerator x over denominator open vertical bar x close vertical bar end fraction rightwards double arrow y equals fraction numerator open vertical bar x close vertical bar open square brackets x close square brackets minus x over denominator open vertical bar x close vertical bar end fraction T h u s comma space y space c a n space a s s u m e space r e a l space v a l u e s space e x c e p t space w h e n space open vertical bar x close vertical bar equals 0 space a n d space open vertical bar x close vertical bar open square brackets x close square brackets minus x equals 0 T h u s comma space y space c a n space a s s u m e space r e a l space v a l u e s space w h e n space open vertical bar x close vertical bar not equal to 0 space a n d space open vertical bar x close vertical bar open square brackets x close square brackets minus x not equal to 0 T h u s comma space y space c a n space a s s u m e space r e a l space v a l u e s space w h e n space open vertical bar x close vertical bar not equal to 0 space a n d space open vertical bar x close vertical bar open square brackets x close square brackets not equal to x T h u s comma space y space c a n space a s s u m e space r e a l space v a l u e s space w h e n space open vertical bar x close vertical bar not equal to 0 space a n d space open vertical bar x close vertical bar less than 1 T h u s space y space c a n space a s s u m e space r e a l space v a l u e s space w h e n space open vertical bar x close vertical bar not equal to 0 space a n d space x element of open parentheses minus infinity comma minus 1 close parentheses union open parentheses 1 comma infinity close parentheses
The following graph depicts the required function.

Answered by Vimala Ramamurthy | 31st May, 2014, 01:06: PM