3.4 Rational function  (Page 6/6)

Problem : Find the range of function :

$$f\left(x\right)=\frac{x}{1+x^2}$$

Solution : The denominator of the given rational function can not be zero. Hence, domain of function is real number set R. There is no exclusion point. Rearranging to form a quadratic equation in x, we have :

$$\Rightarrow y+y{x}^2=x$$

$$\Rightarrow y{x}^2-x+y=0$$

We should analyze for coefficient of “ $x^2$ ” in the quadratic equation. For quadratic equation, coefficient of “ $x^2$ ” can not be zero i.e. y ≠ 0. For real x, y ≠ 0 and D≥0 :

$$D={\left(-1\right)}^2-4XyXy=1-4y^2\ge 0$$ $$\Rightarrow y^2\ge \frac{1}{4}$$ $$\Rightarrow y\in [-\frac{1}{2},\frac{1}{2}]$$

What if y=0? Putting this value in the quadratic equation, we have :

$$\Rightarrow 0-x+0=0$$

$$\Rightarrow x=0$$

This is included in the domain. Hence, y=0 is included in the range. The range of the rational function, therefore, remains unaffected :

$$\Rightarrow y\in [-\frac{1}{2},\frac{1}{2}]$$

Problem : Find the range of the function :

$$y=f\left(x\right)=\frac{x^2-5x+4}{x^2-3x+2}$$

Solution : We see that discrimanants of numerator and denominator polynomials are positive. On factorizing,

$$\Rightarrow y=\frac{x^2-5x+4}{x^2-3x+2}=\frac{\left(x-1\right)\left(x-4\right)}{\left(x-1\right)\left(x-2\right)}$$

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 (x-1) from numerator and denominator :

$$\Rightarrow y=\frac{\left(x-4\right)}{\left(x-2\right)}$$

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 :

$$\Rightarrow y{x}^2-3yx+2y=x^2-5x+4$$ $$\Rightarrow \left(y-1\right)x^2+\left(5-3y\right)x+2y-4=0$$

We should analyze for coefficient of “ $x^2$ ” in the quadratic equation. For quadratic equation, coefficient of “ $x^2$ ” can not be zero i.e. y-1 ≠ 0. For real x, y-1 ≠ 0 and D≥0.

For y-1 = 0, y = 1. Putting this value in the quadratic equation,

$$\Rightarrow 0+\left(5-3\right)x+2-4=0$$ $$\Rightarrow x=1$$

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,

$$D={\left(5-3y\right)}^2-4\left(y-1\right)\left(2y-4\right)\ge 0$$ $$\Rightarrow 25+9y^2-30y-4\{2y^2-6y+4\}\ge 0$$ $$\Rightarrow 25+9y^2-30y-4\{2y^2-6y+4\}\ge 0$$

The coefficient of $$y^2$$ 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 :

$$\Rightarrow y=\frac{\left(x-4\right)}{\left(x-2\right)}$$

Solving,

$$\Rightarrow x=\frac{2y-4}{\left(y-1\right)}$$

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

Problem : Draw the graph of rational function given by :

$$f\left(x\right)=\frac{x^2-1}{x-1}$$

Discuss the nature of graph and also determine domain and range of the given function.

Solution : The form of the given function is that of rational function. We observe that the function is not defined for "x = 1" as function has the form " $x/0$ ", which is undefined. The domain of the given function, therefore, is “R” except “1”. It should be noted that while interpreting domain or range we should not cancel out common terms in the numerator and denominator.

For other values of “x”, the value of the function is given by the reduced expression :

$$f\left(x\right)=\frac{x^2-1}{x-1}=x+1$$

Clearly, if the given function were valid for x =1, then y = x+1 = 1 + 1 = 2. Thus, function f(x) can take any real value except “2”. Hence, range of the function is "R" except "2". The domain and range of the given function are :

$$Domain=R-\left\{1\right\}$$

$$Range=R-\left\{2\right\}$$

In order to plot the function, we calculate few initial values as :

$$Forx=-\mathrm{3,}y=-2$$

$$Forx=-\mathrm{2,}y=-1$$

$$Forx=-\mathrm{1,}y=0$$

$$Forx=\mathrm{0,}y=1$$

$$Forx=\mathrm{1,}\text{y is not defined}$$

$$Forx=\mathrm{2,}y=3$$

$$Forx=\mathrm{3,}y=4$$

The graph of the function is shown here :

The plot is not defined at x = 1. There is a break at x = 1.