10.9

Hello, and welcome back to a step in practice. We’re dealing today with inequalities that involve absolute values. Quite a delicate point. In exercise one, we are asked to solve the inequality absolute value of x to the cube strictly greater than x to the fourth. The domain of the inequality is the set of real numbers. No problem. It is defined everywhere. We know, as Francis showed in the last step, that this is equivalent to x to the cube strictly greater than x to the fourth or minus x to the cube strictly greater than x to the fourth. Now, let me point a few facts. First of all, we consider two inequalities. How do we get them?

64.5

Well, the first inequality is exactly the same, the initial one without the absolute value. And in the second one, there is a minus sign before x to the cube. Why? Well, it’s because the absolute value of x to the cube is either x to the cube or minus x to the cube. And now, how to remind that here, we have to write or and not and. Well, we’re asked to look to the solutions to absolute value of something greater than something else. So we want the absolute value to be big. And whenever we have two sets, the biggest between their union and their intersection is their union. And so we take the or.

111.3

Now, these two inequalities are x to the cube times 1 minus x strictly positive or x to the cube times 1 plus x strictly negative. Let us call this the first inequality, and this the second inequality. We study the first inequality. Well, we can make a table of the values of x to the cube and 1 minus x. So let us write here x. The values that count here are 0 and 1, because x to the cube vanishes at 0, whereas 1 minus x vanishes at 1.

161.3

And let us take here the product x to the cube times 1 minus x. Now let us study the signs. Well, x to the cube vanishes at 0, and it is positive here and negative there. 1 minus x vanishes at 1. Be careful. There is a minus before x, so we reverse the order, the usual order. And we have a plus sign on the left-hand side and a minus sign on the right-hand side. So the product will be 0 here, negative here, positive here, and negative there. So the product will be strictly positive just between 0 and 1. So the first solution is the interval 0, 1. Now let us consider the second inequality.

215.6

Let us draw a table with x. And yet again, we have x to the cube and here 1 plus x. Now, the values that count are minus 1 and 0. So let us write here minus 1, and let us write here 0. So the sign of x to the cube is 0 there, plus, minus, and minus. 1 plus x vanishes at minus 1, and it is positive on the right-hand side and negative on the left-hand side. The product will be positive there, 0 here, negative here, 0 and 0, plus after 0. And we want the product to be strictly negative. So we take the interval from minus 1 to 0.

265.8

So the second set of solutions is the interval minus 1, 0. Now, between the two sets of solutions, we have a union. So we take the union of the two sets, and the solution is the union of the two, which is minus 1, 0 union 0, 1. And this solves exercise 1.

291.2

In exercise two, we are asked to solve the inequality absolute value of x plus 1 strictly greater than absolute value of x plus 2. So one way could be that– consider different cases, depending whether x plus 1 is positive or negative, x plus 2 is positive or negative. Alternatively, you can remark that these are positive quantities. So this inequality is equivalent to the squared inequality, x plus 1 to the square strictly greater than x plus 2 to the square, which means x square plus 2x plus 1 strictly greater than x square plus 4x plus 4, and which gives 2x plus 1 strictly greater than 4x plus 4. And this is equivalent to 2x strictly less than minus 3.

348.7

So x strictly less than minus 3/2. So the set of solutions is the interval from minus infinity to minus 3/2. And this ends exercise two.

364.8

In exercise three, we are asked to solve the following equality– absolute value of x times x minus 2 strictly less than 1. Well, clearly this is equivalent to the fact that x times x minus 2 is between minus 1 and 1. Alternatively, one can write that this is equivalent to the fact that x times x minus 2 is strictly less than 1, and minus x time x minus 2 is strictly less than 1. A way to remind this is that, well, here we have the same inequality without the absolute value. In the second case, we have the inequality without the absolute value and the minus sign inside. Why the minus sign?

422.4

Because the absolute value of a equals a or minus a. Now, why should we take the intersection and not the union? Well, because we are asked to solve an inequality of the form absolute value of something smaller than something else. So we want a small set of solution. And the smallest between the union of two sets and the intersection of the two sets is their intersection. And so here we write and. In any case, we’ve got two inequalities that have to be satisfied at the same time. Now, let us write the first as x to the square minus 2x minus 1 strictly less than 0. And the second is x squared minus 2x plus 1 strictly greater than 0.

473.2

Let us call this the second inequality and this the first inequality. So we study the first inequality. The discriminant is 2 to the square, so 4 plus 4, which is 8, and which is also 2 square root of 2 to the square. So the solutions, the roots of the polynomial, are 2 plus or minus 2 times square root of 2 over 2, which is 1 plus minus square root of 2. And thus, the first inequality, since the coefficient 2 of x square is strictly positive, the polynomial second degree is strictly negative between the two roots. So the solution is the interval 1 minus square root of 2, 1 plus square root of 2.

529.1

And we do not take the roots because we want a strictly negative sign. The second inequality– we can study the discriminant of the polynomial. But we also notice that x square minus 2x plus 1 equals x minus 1 to the square. So it is a strictly positive just whenever it is not 0. So if and only if x is different from 1. So the second set of solutions is the set of real numbers except 1. Now we have to intersect the two sets. Here there is an end. So the intersection, well, is the interval between 1 minus square root of 2, 1 plus square root of 2, except 1.

573.2

So let us write here the solution is the interval 1 minus square root of 2, 1 plus square root of 2 except 1. And this ends exercise three, and also this step. See you in the next step. Bye.