ADDMATHS help???

Question

f(x) = 6/ax+b
f(1) = -6
f(0)=-2
find the value of x such that the function f(x) is not defined

Answer 1

6/a+b=-6 when x=1
6/b=-2 when x=0

therefore b=6/-2=-3
subtitute for b in the first equation
6/a-3=-6
-1=a-3
a=2
now for f(x) subtitute for both b and a
f(x)=6/2x-3
the function is undefined when the denominater is equal to zero
so the domain is for all real numbers except for x=3/2

Answer 2

x = 0

Answer 3

f(1) = -6
6/a(1)+b = 6
6/a+b = 6
6 = 6(a+b)
6a+6b = 6 -------(1)

f(0) = -2
6/a(0)+b = -2
6/b = -2
-2b = 6
b = -3 -------(2)

From (1),
6a+6(-3) = 6
6a-18 = 6
6a = 24
a = 4

Answer 4

f(x)=6/(ax+b)---------------(+)
f(1)=6/(a+b)= -6
-6=6a+6b-------------------1

f(0) = 6/(0+b)
-2 = 6/(b)
-2b = 6
b = -3--------------------------2
Subst.(2) into (1) u have
-6 = 6a - 18
12 = 6a
a = 2

For f(x) to be undefined,the denominator in the function should be ZERO.Thus the sum of ax+b MUST equal to ZERO.

ax+b=0
x = -b/a
Thus x =-(-3)/2 = 3/2------------********
Which will be x = 3/2

Subst(*) into (+) and u have your answer,Will leave that step to you.

Answer 5

f(x) = 6/(ax + b)
f(1) = -- 6
=> 6/(a + b) = -- 6
=> a + b = -- 1 ..............................................(1)
f(0) = -- 2
=> 6/(2a + b) = -- 2
=> 2a + b = -- 3 ............................................(2)
subtracting (1) from (2)
a = -- 2
then b = -- 1 - a = -- 1 + 2 = 1
then f(x) = 6/(1 -- 2x) not defined when 1 -- 2x = 0 or x = 1/2

Answer 6

f(1) = -6
6/a(1) + b = -6 ------------(1)
f(0) = -2
6/a(0) + b = -2
b = -2-----------------------(2)

(2) -> (1)

6/a -2 = -6
a = -6/4
= -3/2

so,
6(-2/3)x -2 = 0
x = -1/2

How am I doing? Is it correct....?

Answer 7

- 6 = 6/(a + b)
- 2 = 6/b
b = - 3
a = 2
f(3/2) is undefined