Solve each rational inequality in Exercises 43–60 and graph the s...

Question

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (x+4)(x−1)/(x+2)≤0

Accepted Answer

Hello today we're going to be solving the given rational inequality. So what we are given is X plus six times X minus four over X plus one. And that should be less than or equal to zero. Now, whenever you want to solve a rational inequality, the first thing you want to do is figure out the value of X that causes the rational function to become undefined. In order to do that. You take the denominator of the rational function which in this case is X plus one and you set it equal to zero and once you saw for X this is going to be the restriction for the given function. So if we subtract one to both sides of this equation, we get X cannot equal to negative one and this is going to be the restriction for X. That causes the inequality to become undefined. Now what we need to do is figure out the zeros for X. In order to plot it on a number line, in order to get the zeros for X, we take the denominator And set each factor of the denominator equal to zero. So doing this is going to give us X plus six is equal to zero and x minus four is equal to zero on the left hand side. If I subtract six to both sides, I get X is equal to negative six. And on the right hand side, if I add four to both sides, I'm going to get X is equal to positive four and these are going to be the zeros for the given inequality. So now that we have the zeros and we have the restrictions, let's go ahead and plot them on the number line. So we're going to have this number line which starts a negative infinity and is going to go to our first zero which is negative six and it's going to go to our next zero which is positive four and end at positive infinity. And keep in mind we also want to plot the restrictions of X on this number line and the only restriction of X that we have is going to be negative one. Since this is a value that causes the inequality to be undefined, we don't want to include it in our solution set. So we're going to assign it an open circle. And for the zeros since the original inequality uses the less than or equal to sign, we're going to be including the values of negative six and four in our solution set. So we're going to represent those with solid dots. Now, what we want to do is take a testing point from the four given regions on the number line, plug it into our original inequality and whichever inequality comes up to be true is going to be the region that we include in our solution set. So let's pick a testing point to the left of -6. This testing point can be when X is equal to negative seven and when x is equal to negative seven, we get negative seven plus six times the quantity negative seven minus four over negative seven plus one less than or equal to zero, negative seven plus six is going to give us negative one and negative seven minus four is going to give us negative 11 and this is going to be over negative seven plus one which is going to give us negative six. Now negative one times negative 11 is going to give us positive 11 and 11 over negative six is going to give us negative 11/ which should be less than equal to zero, which is a true inequality. So since the value of negative seven made the inequality true, we're going to include the region that's to the left of negative six. Now let's pick a testing point between negative six and negative one. And this testing point can be when X is equal to -2. When x is equal to negative two, we get negative two plus six times negative two minus four over negative two plus one less than or equal to zero, negative two plus six is going to give us negative four. I'm sorry, positive four And native to -4 is going to give us -6 and negative two plus one is going to give us negative one which is going to be less than or equal to zero. Four times negative six is going to give us negative 24 negative Over negative one is going to give us positive 24 which is less than equal to zero, which is a false inequality. So since that inequality came out to be false, we're not gonna include any values in the region between -6 and -1. Now we pick a testing point between negative one and four which in this case can be when X is equal to zero. When X is equal to zero, we get zero plus six which is going to give us six times zero minus four, which is going to give us negative 4/0 plus one which is going to give us one and that's going to be less than equal to +06 times negative four is negative 24 negative 24 divided by one is going to give us negative 24 less than equal to zero which is true. So since that came out to be true, we're going to include the region between negative one and positive four. And now we're gonna pick a testing point in the region to the right of positive four. And this can be when x is equal to five. When X is equal to five we get five plus six which is going to be times five minus four which is going to give us one all over five plus one which is going to give us six and this should be less than equal to zero, 11 times one is going to give us positive 11 11/6 is going to give us 11/6. Less than equal to zero, which is a false inequality. So we're not including any values to the right of positive four. Now that we figured out the region's too including our solution said let's now express the solution set interval notation. So the solutions that is going to start from negative infinity and end at -6. So we're starting this first region at negative infinity and we're gonna give negative infinity a parentheses since it doesn't have a finite value and the first region is going to end at negative six. And since negative six is included in the solution set, we're going to assign it a bracket. Then we're going to union the next region which is going to start at one and end F four. And since this region starts at negative one, but doesn't include negative one, we're going to write negative one. But with the parentheses, Then this region is going to end at four. And since four is included, we're going to assign it a bracket. And this is going to be the solution to the rational inequality. And with that being said, the answer to this problem is going to be d. So I hope this video helps you in understanding how to approach this problem and I'll go ahead and see you all in the next video

Solve each rational inequality in Exercises 43–60 and graph the solution set on a real number line. Express each solution set in interval notation. (x+4)(x−1)/(x+2)≤0

Key Concepts

Rational Inequalities

Interval Notation

Graphing Solution Sets

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