Problem :
Find the range of the function :
y
=
f
x
=
x
2
−
5
x
+
4
x
2
−
3
x
+
2
y
=
f
x
=
x
2
−
5
x
+
4
x
2
−
3
x
+
2
Solution :
We see that discrimanants of numerator and denominator polynomials are positive. On factorizing,
⇒
y
=
x
2
−
5
x
+
4
x
2
−
3
x
+
2
=
x
−
1
x
−
4
x
−
1
x
−
2
⇒
y
=
x
2
−
5
x
+
4
x
2
−
3
x
+
2
=
x
−
1
x
−
4
x
−
1
x
−
2
Clearly, rational function is not defined for x=1 and x=2. Domain of the function is R {1,2). For the sake of determining range, the limiting values of function for these values of x are obtained by canceling (x1) from numerator and denominator :
⇒
y
=
x
−
4
x
−
2
⇒
y
=
x
−
4
x
−
2
For x=1, y = 3. For x=2, however, the function value is indeterminate. In totality, we need to exclude y=3 from the interval of real values of y. Now, in order to determine real values of y, we rearrange the given function to form a quadratic equation in x :
⇒
y
x
2
−
3
y
x
+
2
y
=
x
2
−
5
x
+
4
⇒
y
x
2
−
3
y
x
+
2
y
=
x
2
−
5
x
+
4
⇒
y
−
1
x
2
+
5
−
3
y
x
+
2
y
−
4
=
0
⇒
y
−
1
x
2
+
5
−
3
y
x
+
2
y
−
4
=
0
We should analyze for coefficient of “
x
2
x
2
” in the quadratic equation. For quadratic equation, coefficient of “
x
2
x
2
” can not be zero i.e. y1 ≠ 0. For real x, y1 ≠ 0 and D≥0.
For y1 = 0, y = 1. Putting this value in the quadratic equation,
⇒
0
+
5
−
3
x
+
2
−
4
=
0
⇒
0
+
5
−
3
x
+
2
−
4
=
0
⇒
x
=
1
⇒
x
=
1
We see that x=1 is not part of domain. This is actually the value which reduces denominator to zero. Hence, we should exclude y = 1 from the real values of y. Now for D≥0,
D
=
5
−
3
y
2
−
4
y
−
1
2
y
−
4
≥
0
D
=
5
−
3
y
2
−
4
y
−
1
2
y
−
4
≥
0
⇒
25
+
9
y
2
−
30
y
−
4
{
2
y
2
−
6
y
+
4
}
≥
0
⇒
25
+
9
y
2
−
30
y
−
4
{
2
y
2
−
6
y
+
4
}
≥
0
⇒
25
+
9
y
2
−
30
y
−
4
{
2
y
2
−
6
y
+
4
}
≥
0
⇒
25
+
9
y
2
−
30
y
−
4
{
2
y
2
−
6
y
+
4
}
≥
0
The coefficient of
y
2
y
2
is positive. The discriminant is 0. Clearly, following sign rule, f(x) ≥0 for all real values of y. Hence, real values of y are real number set R. However, we need to exclude y = {1,3) as discussed above. Therefore, range of given function is R{1,3}.
Alternative
Once, exception points are noted, we can evaluate “y” from the reduced form :
⇒
y
=
x
−
4
x
−
2
⇒
y
=
x
−
4
x
−
2
Solving,
⇒
x
=
2
y

4
y
−
1
⇒
x
=
2
y

4
y
−
1
Clearly, y#1. But we have seen that y#3 as well. Hence, range of rational function is R{1,3}.