give information about the following equation?

Question

f: R --> R
f(x) = (1/2) * (x + IxI)
a) is f 1-1 and onto? Justify your answer.
b) What is the range of f on a) R and b) R+

Answer 1

x + |x| maps negative numbers on 0 and all other on x

so
a) NO see my observations above.
b) Range a) R+and 0, b)R+

Tell me have you learned something from me ???

Answer 2

function rewritten:

f(x)=1/2*(x+IxI)

if x=0 f(x)=1
for every x>0, x+ IxI=2x=> f(x)=1/2*2x=2*(-2x)>0
for every x<0, x+ IxI= 0=> f(x)=1

in an 1-1 function, two elements of Domain must be preimages of two different elements of the co-domain. that means, f(a)=f(a')
if and only if a=a'

in this case, f(0)=f(-1)=1, thus it is not 1-1.

in an onto function, Range equates Co-domain. here
Range=(0,1]
Co-Domain=(-α,α)

thus, it is not an onto function too.

b) for R the range will be (0,1]
for R+ the range will be (0,1)

Answer 3

a) no it is not 1-1 onto
b)range on R all real numbers from 0 to infinity
on R+ the range is all positive real numbers

Answer 4

f is not 1-1 as for -ve x we always get a 0.
It is not onto as well as for -ve integers in range R we have no mapping from domain(R)
Range is [0,inf) for R and [1,inf) for R+

Answer 5

a) It's not 1-1 as if x=a as a>0
f(x)=a
if x=-a as a>0
f(x)=0
so any negative value of x lead f(x)=0
so its R+ -->R so its not onto
b) f range is from zero to infinity so its R+(if you consider R+ contain zero, some bodies doesnpt consider that )