3.4 Rational function  (Page 5/6)

Problem : Find value of x for which given function has least value.

$$y=f\left(x\right)=\frac{x^2-6x+5}{x^2+2x+1}$$

Solution : Rearranging to form a quadratic equation in x, we have :

$$\Rightarrow y{x}^2+2yx+y=x^2-6x+5$$ $$\Rightarrow \left(y-1\right)x^2+2\left(y+3\right)x+y-5=0$$

For x real, D≥0.

$$\Rightarrow 4{\left(y+3\right)}^2-4\left(y-1\right)X\left(y-5\right)\ge 0$$ $$\Rightarrow \left(y^2+6y+9\right)-\left(y^2-6y+5\right)\ge 0$$ $$\Rightarrow 12y+4\ge 0$$ $$\Rightarrow y\ge -\frac{1}{3}$$

The real values of y, therefore, lies in the interval [-1/3, ∞). The least value of y in the interval of real values is -1/3. We should, however, check that value of y=-1/3 does not correspond to value of x which is not permitted. Here, the denominator polynomial is $x^2+2x+1={\left(x+1\right)}^2$ . Thus, $x\ne -1$ . The domain of the function is R-{-1}.

Now, we calculate value of x corresponding to y as :

$$\Rightarrow -\frac{1}{3}=\frac{x^2-6x+5}{x^2+2x+1}$$

$$\Rightarrow x^2-4x+4=0$$ $$\Rightarrow {\left(x-2\right)}^2=0$$ $$\Rightarrow x=2$$

This point belongs to the domain of the function as it is different to excluded value. Hence, least value of y is -1/3.

Problem : For what values of “a”, the function given here assumes all real values for real x.

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

Solution : Rearranging to form a quadratic equation in x, we have :

$$\Rightarrow 3yx-4y{x}^2+ay=a{x}^2+3x-4$$

$$\Rightarrow \left(a+4y\right)x^2+3\left(1-y\right)x-\left(ay+4\right)=0$$

For x real, D≥0.

$$\Rightarrow 9{\left(1-y\right)}^2+4\left(a+4y\right)X\left(ay+4\right)\ge 0$$

$$\Rightarrow \left(9+16a\right)y^2+\left(4a^2+46\right)y+\left(9+16a\right)\ge 0$$

This is a quadratic inequality in y. This inequality holds when coefficient of “ $y^2$ ” is positive and discriminant is less than equal to zero i.e. D’≤0.

$$\Rightarrow \left(9+16a\right)>0$$

$$\Rightarrow a>-\frac{9}{16}$$

and

$$D\prime ={\left(4a^2+46\right)}^2-4{\left(9+16a\right)}^2\le 0$$ $$\Rightarrow \left(2a^2+23\right){}^2-{\left(9+16a\right)}^2\le 0$$ $$\Rightarrow \left(2a^2+23+9+16a\right)\left(2a^2+23-9-16a\right)\le 0$$ $$\Rightarrow \left(2a^2+16a+32\right)\left(2a^2-16a+14\right)\le 0$$ $$\Rightarrow \left(a^2+8a+16\right)\left(a^2-8a+7\right)\le 0$$ $$\Rightarrow {\left(a+4\right)}^2\left(a^2-8a+7\right)\le 0$$ $$\Rightarrow \left(a^2-8a+7\right)\le 0$$ $$\Rightarrow \left(a^2-8a+7\right)\le 0$$

The roots of corresponding quadratic equation in a is :

$$\Rightarrow a=\mathrm{1,7}$$

Hence interval of a satisfying inequality is [1,7]

We have two intervals i.e. a>-9/16 and $\Rightarrow a\in \left[\mathrm{1,7}\right]$ corresponding to two simultaneous conditions. Therefore, values of a is intersection of these two intervals :

$$\Rightarrow a\in \left[\mathrm{1,7}\right]$$