Answer:
Answer:
D=(-\infty,3)\cup(3,\infty) [x|x\neq3]D=(−∞,3)∪(3,∞)[x∣x
=3]
R=(-\infty,-1)\cup(-1,\infty) [y|y\neq-1]R=(−∞,−1)∪(−1,∞)[y∣y
=−1]
Given : f(x)=\frac{x-2}{3-x}f(x)=
3−x
x−2
To find : The domain and range of the real function
Solution :
To find domain : Equate the denominator to zero
f(x)=\frac{x-2}{3-x}f(x)=
3−x
x−2
Denominator (3-x)=0
x=3
This means at x=3 function is not defined
And by definition of domain - The domain is where the function is not defined.
Domain is D=(-\infty,3)\cup(3,\infty) [x|x\neq3]D=(−∞,3)∪(3,∞)[x∣x
=3]
Range
Put f(x)=y
y=\frac{x-2}{3-x}y=
3−x
x−2
x=\frac{3y-2}{y+1}x=
y+1
3y−2
Range is the set of value that correspond to domain
Equate the denominator to zero
x=\frac{3y-2}{y+1}x=
y+1
3y−2
Denominator (y+1)=0
y=-1
This means at y=-1 function is not defined
Range is R=(-\infty,-1)\cup(-1,\infty) [y|y\neq-1]R=(−∞,−1)∪(−1,∞)[y∣y
=−1]
