Question

The function f: R--->R defined as \[f(x) =\frac{ 3x^2 + 3x - 4 }{ -4x^2 + 3x + 3 }\] check wheather it is one one or on to funtions

Answer 1

have you graphed it?

Answer 2

Nope..)

Answer 3

(4y+3)x2+(3-3y)x-(4+3y)=0
check delta if delta>0 it is not one to one

Answer 4

if for all y delta<=0 so it is one to one

Answer 5

delta=b2-4ac

Answer 6

because for a y it gives you two x for some y.

Answer 7

delta=(3-y)2+(4+3y)2 >0
so it is not one to one.

Answer 8

yup....@mahmit2012 that make lot of sense....then hw to check for onto condition..?

Answer 9

for all y#-3/4 you have some x for y=-3/4 check .you have also x=0

Answer 10

so it is onto also

Answer 11

not zero but there is a x

Answer 12

BUT the codomain is R, not R\{-3/4} so it is NOT onto since y=-3/4 has no pre-image x.

Answer 13

why I checked y=-3/4
because it is x2 coefficent or horizental asymtotic line

Answer 14

@Zarkon right, but for your stage and second order equation that's better to figure out with delta.

Answer 15

sorry, my mistake.

Answer 16

but strictly speaking, the relation is not a function since the relation is not well defined, i.e., there are two values of x that makes the denominator zero.

Answer 17

y=-3/4-> 21x/9=-7/4
x=-3/4 check it out!

Answer 18

if both coefficient of x and x2 were the same so you right, but it is not @sirm3d

Answer 19

the question states that "f: R--->R"

which is not true

Answer 20

@mahmit2012 i did nt get u ( onto) ?

Answer 21

onto means for all y you have at least one x.
so the second equation tells you it is.

Answer 22

wat is second eq here

Answer 23

@Zarkon That is related to domain not for function to be onto.

Answer 24

f:A->B for checking onto you should pay attention to B and cover B in Domain belonging to A.

Answer 25

the function is ill defined if they write "f:R->R"

Answer 26

they should write f:A->R
where A is the maximal set in which f can be defined