Let f(x) = ( 3x-4)/(5x+2) If the range of f(x) is Z ( the integers) then what is the domain.?

All values of f(x) are integers, so the ratio (3x-4)/(5x+2) is always an integer. Therefore 3x-4 = k(5x+2), where k is an integer.

Solving for x, we get x = (-2k-4)/(5k-3). Since k is an integer, so are 2k, 5k, -2k-4, and 5k-3. Therefore x is a rational number, whose only disallowed value is 3/5 (which would cause a zero denominator).

In set notation, the domain is {x | x ∈ Q, x ≠ 3/5}.

---edited to fix a sign error---

f(x)=(3x-4)/(5x+2), f(x) must be an integer.

(5x+2)f(x)=3x-4

=>

5xf(x)+2f(x)=3x-4

=>

x=-2[f(x)+2]/[5f(x)-3]

=>

f(x)|.. -4......-3...-2...-1...0..

..x.| -4/23..-1/9..0..1/4..4/3

f(x)|...1......2.....3.......4....

..x.|..-3..-8/7..-5/6..-12/17

From this table, we see that

f(x)=3/5 is impossible, for then

f(x) is not an integer. Besides,

there are infinitely many real x

which will not be an element of

the domain of f(x). We can only

write roughly

Dom(f)={...-4/23,-1/9,0,1/4,4/3,

-3,-8/7,-5/6,-12/17....}

no general rule can be found.

"argent ' has the correct concepts but D = { all rationals of the form ( 2 k +4 ) / ( 3 - 5 k ) , where k = integer }...NOTE : 3 - 5k is never 0

.........( 3x-4)

f(x) = --------

.........(5x+2)

// You cannot have a zero in the denominator,

// therefore, 5x+2≠0

5x+2 ≠ 0 ⇒

5x ≠ -2 ⇒

x ≠ -⅖

Domain = all x in ℝ such that x ≠ -⅖, written (-∞, -⅖) U (⅖, +∞).............ANS

Domain is x where 5x +2 does not = 0

x =/= -0.4