# 3.5 Rational inequality  (Page 6/6)

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$x={y}^{n}$

Interchangeably, we write :

$y={x}^{\frac{1}{n}}={}_{\sqrt{}}^{n}\left(x\right)$

If n is even integer, then x can not be negative. For n=2, we drop “n” from the notation and we write,

$y=\sqrt{x}$

We extend this concept to function in which number “x” is substituted by any valid expression (algebraic, trigonometric, logaritmic etc). Some examples are :

$y=\sqrt{{x}^{2}+3x-5}$ $y=\sqrt{\mathrm{log}{}_{e}\left({x}^{2}+3x-5\right)}$

We shall also include study of radical function which is part of rational form like :

$y=\sqrt{\left(\frac{1}{{x}_{2}+3x-5}\right)}$

$y=\sqrt{\left\{\frac{\left(x-1\right)\left(x+3\right)}{{x}_{2}+3x-5}\right\}}$

Analysis of root function is same as analysis of inequality of function. Because, radical function ultimately results in inequality. We make use of the fact that expression within the radical sign is non-negative. Here, we denote a radical function as :

$f\left(x\right)=\sqrt{g\left(x\right)}$

As the expression under is non-negative,

$⇒g\left(x\right)\ge 0$

When radical function is part of a function defined in rational form, the radical function should not be zero. Let us consider a function as :

$f\left(x\right)=\frac{1}{\sqrt{g\left(x\right)}}$

As the radical is denominator of the rational expression, expression under radical sign is positive,

$g\left(x\right)>0$

We have already worked with inequalities involving polynomial and rational functions. We shall restrict ourselves to few illustrations here.

Problem : Find domain of the function :

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

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

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

$⇒1-{x}^{2}\le 1$ $⇒{x}^{2}\ge 0$

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

$⇒1-{x}^{2}\ge 0$ $⇒\left(1+x\right)\left(1-x\right)\ge 0$

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

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

Using sign rule :

$⇒x\in \left[-1,1\right]$

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

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

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.

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

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, ${x}^{a}>{x}^{b},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}a>b$ . 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, ${x}^{a}<{x}^{b},\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{1em}{0ex}}a>b$ . Rearranging in groups,

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

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

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

3: $x<0$

Rearranging in groups,

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

Here, ${x}^{14,}{x}^{6}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{2}$ are positive and ${x}^{11}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{3}$ is negative. Hence, total expression is positive,

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

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

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

## Exercise

Find solution of the rational inequality given by :

$\frac{\left(x+1\right)\left(x+5\right)}{\left(x-3\right)}\ge 0$

Hint : Critical points are -5,-1 and 3. We need to exclude end corresponding to x=3 as denominator turns zero for this value.

$\left[-5,-1\right]\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}\left(3,\infty \right)$

Find solution of the rational inequality given by :

$\frac{8{x}^{2}+16x-51}{\left(2x-3\right)\left(x+4\right)}>0$

Hint : Critical points are -4,-3,3/2,5/2.

$⇒x\in \left(-\infty ,-4\right)\cup \left(-3,3/2\right)\cup \left(5/2,\infty \right)$

Find solution of the rational inequality given by :

$\frac{{x}^{2}+4x+3}{{x}^{3}-6{x}^{2}+11x-6}>0$

Hint : Factorize denominator as ${x}^{3}-6{x}^{2}+11x-6=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ . Critical points are -3,-1,1,2,3.

$⇒x\in \left(-3,-1\right)\cup \left(1,2\right)\cup \left(3,\infty \right)$

Find solution of the rational inequality given by :

$\frac{\left(2x+1\right){\left(x-1\right)}^{2}}{\left({x}^{3}-3{x}^{2}+2x\right)}\ge 0$

Hint : Factorize denominator as

$⇒\frac{\left(2x+1\right){\left(x-1\right)}^{2}}{\left({x}^{3}-3{x}^{2}+2x\right)}=\frac{\left(2x+1\right){\left(x-1\right)}^{2}}{x\left(x-1\right)\left(x-2\right)}$

Critical points are -1/2,1,1,0,1 and 2. We see that "1" is repeated odd times. Hence, we continue to assign alternating signs in accordance with wavy curve method. The solution of x for the inequality is :

$-\infty

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