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**Rational Inequalities**

Solving Rational Inequalities Rational Inequalities Solving Rational Inequalities Mathematics can be both abstract and applied. Abstract mathematics is focused on axioms, theorems, and proofs. It can be derived independently of empirical evidence. Theorems that were proved centuries ago are still valid today. In this sense, abstract mathematics transcends time. Yet, even though mathematics can be developed in an abstract setting – separate from science and all measured data – it also has countless applications. There is a common misconception that theoretical mathematics is unimportant, yet many of the ideas that eventually had great practical importance were first born in the abstract. For example, in 1854 George Boole published Laws of Thought, which outlined the basis for Boolean algebra. This was 85 years before the invention of the first digital computer. However, Boolean algebra became the basis by which modern computer hardware operates. Much like Boolean algebra, the purpose of complex numbers was at first theoretical. However, today complex numbers are used in the design of electrical circuits, ships and airplanes. Basic quantum physics and certain features of the theory of relativity would not have been developed without complex numbers. Even fractal images [ widely used in computer graphics ] would not have been discovered without complex numbers. In this chapter we discuss some important topics in algebra that have had an impact on society. We are privileged to read in a few hours what took people centuries to discover. To ignore either the abstract beauty or the profound applicability of mathematics is like seeing a rose but never smelling one. Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities General Form f(x) = where p(x) and q(x) are polynomials General Strategy Solve related equations p(x) = 0, q(x) = 0 to find boundary points Check sign of f(x) between pairs of consecutive boundary points p(x) q(x) > 0 ... or ≥ or < or ≤ Rational Inequalities: Forms and Strategies Here we can no longer just transform the inequality into an equation and solve for the boundary points. We must solve equations for each of the numerator and denominator to find the boundary points and asymptotes. Finding the zeros of the numerator polynomial shows us where the zeros are located – that is, boundary points for solution intervals. Finding the zeros of the denominator polynomial shows us where the vertical asymptotes are located – that is, the points where the function fails to exist. These are also boundary points for solution intervals. Dividing the x-axis into intervals bounded by the zeros and asymptotes found above shows us the possible solution intervals. Check the signs of p(x) and q(x) in each of the intervals bounded by the boundary points. These determine the sign of the rational function f(x) in those intervals, and hence which intervals contain solutions for the inequality. 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities x y Solve 1 x2 > Example: Note that exists for all x 1 x2 1 x2 = y except x = 0 Since x2 > 0 for all x then 1 x2 > ∞ = (– ,0) (0, ) ∩ Solution set: { x x ≠ 0 } Rational Inequalities: Example In the first example, we see from both the symbolic form of the function and from its graph that it is positive everywhere that it is defined. However, It is not defined only at x = 0. Hence the solution set is { x | x ≠ 0 } = (-, 0) (0, ) In the second example, we note that the function is positive everywhere defined and does not exist at x = 0. Hence the function can never be less than or equal to 0; that is, the inequality is false for all x in the domain of the function, giving the inequality no solutions. The solution set is { } = ∅. Another technique that could be used in both examples is multiplying by x2 (where x ≠ 0) to yield the inequalities 1 > 0 and 1 ≤ 0, respectively. The first of these is true for all x, except for x = 0 where the function is not defined. The second is true for no x, which exactly matches our solution set. Great care must taken if this technique is to be used in order to draw the correct conclusions. So, the answer to the question posed is YES. Note that we could NOT start by solving 1 x2 = WHY ? 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities x y 1 x2 = Solve 1 x2 ≤ Example: Since x2 > 0 for all x then 1 x2 > Hence there is NO x for which 1 x2 ≤ Solution set: { } Question: Rational Inequalities: Example In the first example, we see from both the symbolic form of the function and from its graph that it is positive everywhere that it is defined. However, It is not defined only at x = 0. Hence the solution set is { x | x ≠ 0 } = (-, 0) (0, ) In the second example, we note that the function is positive everywhere defined and does not exist at x = 0. Hence the function can never be less than or equal to 0; that is, the inequality is false for all x in the domain of the function, giving the inequality no solutions. The solution set is { } = ∅. Another technique that could be used in both examples is multiplying by x2 (where x ≠ 0) to yield the inequalities 1 > 0 and 1 ≤ 0, respectively. The first of these is true for all x, except for x = 0 where the function is not defined. The second is true for no x, which exactly matches our solution set. Great care must taken if this technique is to be used in order to draw the correct conclusions. So, the answer to the question posed is YES. Could we solve either inequality by rewriting ? What if we clear the fraction by writing : ≤ 1 x2 (x2) > 1 x2 (x2) OR ? 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities 2x (x – 2 )2 > Example: Solve Note that x ≠ 2 Since (x – 2)2 ≥ 0 for all x and 2x > 0 provided x ≠ 0 then for all x > 0 except for x = 2 x y 2x (x – 2 )2 > x = 2 Rational Inequalities: Example Here we simply note that the denominator is always non-negative and the numerator is positive only if x is positive. Hence, the rational function is positive only when x is. We also note that, when x = 2 the rational function does not exist. For boundary points we see that the function is 0 when x = 0 and the function fails to exist at x = 2. Hence the inequality holds for all positive x except x = 2. The solution set is thus { x | 0 < x < 2 } { x | 2 < x } = (0, 2) (2, ) Note again that the intervals are open intervals and that cannot ever be included in an interval since it is not a number but only a direction. Solution set: (0, 2) (2, ) ∞ ∩ = { x 0 < x < 2 } { x 2 < x } ∩ 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities 2 – x 3 > x 2 + x Example: Note that x ≠ 2 and x ≠ – 2 Solve x y g(x) = x 2 + x g(x) 3 2 – x f(x) , y = 1 If then for what x is f(x) > g(x) ? f(x) Clearly f(x) > g(x) only for x = –2 x = 2 –2 < x < 2 Rational Inequalities: Example The simplest way to solve this inequality is graphically. By allowing the left side to be called f(x) and the right side g(x) we restate the inequality as the problem of finding all values of x for which f(x) > g(x). Noting that f(x) has a vertical asymptote at x = 2 and horizontal asymptote of the x-axis, that is, the line y = 0, we graph f(x). Then, noting that g(x) has a vertical asymptote at x = –2 and horizontal asymptote of y = 1, we graph g(x). The thing to note now is that for every value of x outside the interval (–2, 2) we have g(x) > f(x) and for x inside the interval (–2, 2) we have f(x) > g(x). Hence, the solution set is the interval (–2, 2) or { x –2 < x < 2 } . Also note that at no place do we have f(x) = g(x). Thus any attempt to solve the “associated” equation must fail, since the graphs never intersect. Attempts to solve symbolically by this method produce only complex numbers for values of x – i.e. there is no real solution. Solution set: { x | –2 < x < 2 } = (–2, 2) Question: Can we solve by solving the related equation for boundary points ? 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities x y Example: 5 x ≥ 2 x2 + f(x) = – Solve 5 x 1 x2 Note: and exist y = 2 ● for all x except x = 0 Assuming x ≠ 0 then 5 x 2 x2 + f(x) = – = 2x2 – 5x + 2 x2 Rational Inequalities: Example The boundary points are the places where we have vertical asymptotes and places where the graph of f(x) crosses the x-axis. The asymptote occurs at x = 0 and the function f(x) = 0 at x = ½ and x = 2. So x = 0 is a boundary point. We test f(x) at the points x = –1 , x = .1 , x = 1, and x = 3 that lie in the intervals formed by the boundary points. We see that f(x) is positive in all the intervals except for (½ , 2). Since the inequality is a non-strict inequality (with ≥), then x = ½ and x = 2 are included as solutions and therefore are boundary points. Hence, the solution set is the union of the intervals shown, also representable as { x x < 0 } { x 0 < x ≤ ½ } { x 2 ≤ x } where the ≤ represents the inclusion of a boundary point, that is, the square bracket ] and [ in interval notation. ≥ For x ≠ 0 denominator is positive So f(x) has sign of the numerator 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities x y y = 2 ● Example: ≥ = 2x2 – 5x + 2 x2 f(x) So f(x) has the sign of the numerator ● ● Set numerator to zero: 2x2 – 5x + 2 = 0 Rational Inequalities: Example The boundary points are the places where we have vertical asymptotes and places where the graph of f(x) crosses the x-axis. The asymptote occurs at x = 0 and the function f(x) = 0 at x = ½ and x = 2. So x = 0 is a boundary point. We test f(x) at the points x = –1 , x = .1 , x = 1, and x = 3 that lie in the intervals formed by the boundary points. We see that f(x) is positive in all the intervals except for (½ , 2). Since the inequality is a non-strict inequality (with ≥), then x = ½ and x = 2 are included as solutions and therefore are boundary points. Hence, the solution set is the union of the intervals shown, also representable as { x x < 0 } { x 0 < x ≤ ½ } { x 2 ≤ x } where the ≤ represents the inclusion of a boundary point, that is, the square bracket ] and [ in interval notation. 1 2 … so boundary points at Solutions: and 2 x = 0, x = , x = 2 1 2 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Rational Inequalities**

Solving Rational Inequalities x y y = 2 Example: ≥ = 2x2 – 5x + 2 x2 f(x) x = 0, x = , x = 2 1 2 ● ● ● Checking intervals: x = –1 f(–1) = 9 > 0 x = 0.1 f(.1) = 2.52 > 0 Rational Inequalities: Example The boundary points are the places where we have vertical asymptotes and places where the graph of f(x) crosses the x-axis. The asymptote occurs at x = 0 and the function f(x) = 0 at x = ½ and x = 2. So x = 0 is a boundary point. We test f(x) at the points x = –1 , x = .1 , x = 1, and x = 3 that lie in the intervals formed by the boundary points. We see that f(x) is positive in all the intervals except for (½ , 2). Since the inequality is a non-strict inequality (with ≥), then x = ½ and x = 2 are included as solutions and therefore are boundary points. Hence, the solution set is the union of the intervals shown, also representable as { x x < 0 } { x 0 < x ≤ ½ } { x 2 ≤ x } where the ≤ represents the inclusion of a boundary point, that is, the square bracket ] and [ in interval notation. x = f(1) = –1 < 0 No Solutions ! x = f(3) = 5 > 0 (– , 0) ∞ [ 2, ) ∩ (0, ) 1 2 Solution set: 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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**Solving Rational Inequalities**

Think about it ! 7/9/2013 Rational Inequalities Rational Inequalities 7/9/2013

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