3.5 Rational inequality  (Page 6/6)

Problem : Find domain of the function :

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

Solution : One radical (inner) is contained with another radical (outer). For the outer radical,

$$\Rightarrow 1-\sqrt{1-x^2}\ge 0$$ $$\Rightarrow \sqrt{1-x^2}\le 1$$

The term on each side of inequality is a positive quantity. Squaring each side does not change inequality,

$$\Rightarrow 1-x^2\le 1$$ $$\Rightarrow x^2\ge 0$$

This quadratic inequality is true for all real x. Now, for inner radical

$$\Rightarrow 1-x^2\ge 0$$ $$\Rightarrow \left(1+x\right)\left(1-x\right)\ge 0$$

We multiply by -1 to change the sign of x in 1-x,

$$\Rightarrow \left(x+1\right)\left(x-1\right)\le 0$$

Using sign rule :

$$\Rightarrow x\in [-\mathrm{1,1}]$$

Since, conditions corresponding to two radicals need to be fulfilled simultaneously, the domain of the given function is intersection of outer and inner radicals.

$$\text{Domain}=[-\mathrm{1,1}]$$

Problem : Find the domain of the function given by :

$$f\left(x\right)=\sqrt{\left(x^{14}-x^{11}+x^6-x^3+x^2+1\right)}$$

Solution : Clearly, function is real for values of “x” for which expression within square root is a non negative number. We note that independent variable is raised to positive integers. The nature of each monomial depends on the value of x and nature of power. If x≥1, then monomial evaluates to higher value for higher power. If x lies between 0 and 1, then monomial evaluates to lower value for higher power. Further, a negative x yields negative value when raised to odd power and positive value when raised to even power. We shall use these properties to evaluate the expression for three different intervals of x.

$$\Rightarrow x^{14}-x^{11}+x^6-x^3+x^2+1\ge 0$$

We consider different intervals of values of expression for different values of “x”, which cover the complete interval of real numbers.

1: $x\ge 1$

In this case, $x^a>x^b,\text{if}a>b$ . Evaluating in groups,

$$\left(x^{14}-x^{11}\right)+\left(x^6-x^3\right)+\left(x^2+1\right)>0$$

2: $0\le x<1$

In this case, $x^a<x^b,\text{if}a>b$ . Rearranging in groups,

$$x^{14}-\{\left(x^{11}-x^6\right)+\left(x^3-x^2\right)\}+1$$

Here, $\{\left(x^{11}-x^6\right)+\left(x^3-x^2\right)\}$ is negative. Hence total expression is positive,

$$\Rightarrow x^{14}-\{\left(x^{11}-x^6\right)+\left(x^3-x^2\right)\}+1>0$$

3: $x<0$

Rearranging in groups,

$$\left(x^{14}-x^{11}\right)+\left(x^6-x^3\right)+\left(x^2+1\right)$$

Here, $x^{\mathrm{14,}}{x}^6\text{and}{x}^2$ are positive and $x^{11}\text{and}{x}^3$ is negative. Hence, total expression is positive,

$$\Rightarrow \left(x^{14}-x^{11}\right)+\left(x^6-x^3\right)+\left(x^2+1\right)>0$$

We see that expression is positive for all values of “x”. Hence, domain of the function is :

$$\text{Domain}=R=\left(-\infty ,\infty \right)$$