Question

If \(\frac{y^{2}+y-6}{y^{2}+y-2}<0\), then which of the following is true? (1) \(1<\mathrm{y}<2\) (2) \(-3<\mathrm{y}<-2\) (3) Either (1) or (2) (4) Neither (1) nor (2)

Short answer

Answer: Option (3) Either (1) or (2).

Step-by-step solution

Factor the numerator and denominator

Factor the given expression: $$ \frac{y^{2}+y-6}{y^{2}+y-2}=\frac{(y-2)(y+3)}{(y-1)(y+2)}. $$ This will allow us to determine the critical points.

Determine critical points

To find the critical points, set each of the factors of the numerator and denominator equal to zero: $$ (y-2)=0, \quad (y+3)=0, \quad (y-1)=0, \quad (y+2)=0. $$ Solving each equation, we find the critical points: \(y = 2,-3,1,-2\).

Test intervals

Now we test the intervals between the critical points to determine which parts satisfy the inequality. Let us define a test function \(f(y)\) as: $$ f(y)=\frac{(y-2)(y+3)}{(y-1)(y+2)}. $$ Test intervals: 1. \(y < -3\): Choose \(y=-4\). \(f(-4)=\frac{(-6)(1)}{(5)(-2)}>0\). 2. \(-3 < y < -2\): Choose \(y=-2.5\). \(f(-2.5)=\frac{(0.5)(0.5)}{(2.5)(0.5)}<0\). 3. \(-2 < y < 1\): Choose \(y=0\). \(f(0)=\frac{(-2)(3)}{(-1)(2)}>0\). 4. \(1 < y < 2\): Choose \(y=1.5\). \(f(1.5)=\frac{(-0.5)(4.5)}{(2.5)(3.5)}<0\). 5. \(y > 2\): Choose \(y=3\). \(f(3)=\frac{(1)(6)}{(7)(5)}>0\).

Identify true option

From the test intervals, we can see that the inequality \(\frac{(y-2)(y+3)}{(y-1)(y+2)}<0\) holds true only for the intervals of \((-3, -2)\) and \((1, 2)\). Hence, either option (1) or (2) must be true. Thus, the correct answer is: Option (3) Either (1) or (2).

Key concepts

Rational Functions

Rational functions are mathematical expressions that consist of a ratio of two polynomials. They are written in the form of \(\frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials. These types of functions are known for having asymptotes and discontinuities, typically where the denominator equals zero and thereby is undefined.

In the given exercise, the rational function is \( \frac{y^{2}+y-6}{y^{2}+y-2} \). Both the numerator and the denominator are quadratic polynomials, meaning that the degree of this rational function is zero (polynomials of the same degree).

Understanding rational functions involves knowing how to manipulate these expressions, which often means factoring them. Factoring helps identify zeros and other critical points that are useful for analysis, especially in solving inequalities.

Solving Inequalities

Solving inequalities with rational functions involves finding the values of the variable that satisfy the inequality condition. Unlike equations, which give specific solutions, inequalities often yield ranges of values.

For a rational function like \( \frac{y^{2}+y-6}{y^{2}+y-2} < 0 \), the first step is typically to factor both the numerator and the denominator, leading to \( \frac{(y-2)(y+3)}{(y-1)(y+2)} < 0 \).

The goal is to determine where the expression is negative. This often involves considering both the numerator and denominator separately to find the sign of each part in given intervals. Any value that makes the denominator zero is an important point to exclude since it will make the function undefined.

Interval Testing

Interval testing is a method often used in solving inequalities with rational functions. It involves dividing the real number line into intervals, determined by the critical points, and then testing points within these intervals to determine the sign of the function.

In this exercise, the critical points obtained were \( y = 2, -3, 1, -2 \). These points divide the number line into several intervals:

• \( y < -3 \)
• \( -3 < y < -2 \)
• \( -2 < y < 1 \)
• \( 1 < y < 2 \)
• \( y > 2 \)

By selecting a test point from each interval and substituting it into the factored rational expression, you can determine the inequality's validity in each interval, whether the expression is positive or negative. The correct intervals are those where the solution to the inequality is true.

Critical Points

Critical points in a rational function are the values of the variable where the function changes its behavior, such as change in sign or becoming undefined. They are found by setting each factor of the numerator and denominator to zero separately.

In our exercise, factoring the expression \( \frac{(y-2)(y+3)}{(y-1)(y+2)} \) provided the solutions for the equations \((y-2)=0\), \((y+3)=0\), \((y-1)=0\), and \((y+2)=0\). Solving these gives the critical points \( y = 2, -3, 1, -2 \).

These points are essential as they are where the function could switch from positive to negative or vice versa. They also mark where the denominator is zero and the function is undefined. Therefore, these points help us segment the number line into different intervals for testing, and thus, identify the valid solutions for a given inequality.