 Hello and welcome to the session. In this session we will discuss Rational and Exponential Inequalities. First of all let us discuss Rational Inequality. Now Rational Inequality is of the type P upon Q is greater than equal to 0. P upon Q is less than equal to 0. P upon Q is greater than 0 and P upon Q is less than 0 where P and Q are non-zero polynomials. Now when right hand side of inequality is 0 it is also called standard form of the inequality. Now Rational Inequalities are solved in the same way as we solved quadratic inequalities. The only difference is in the critical numbers we include zeros of the denominator in our list of critical values but we do not include the critical values given by denominator in the solution set because at that value the denominator will be 0 and the function P upon Q will not be defined at that point thus zeros of the denominator are also called descriptions. Now we will follow some steps to find solution set of rational inequality In step 1 write the given inequality in standard form that is we will make the right hand side of the given inequality equal to 0 then in second step find all zero zeros of polynomials P and Q known as critical values then in next step we will plot all zero zeros of polynomials P and Q on the number line Now in step 4 choose test number in each interval of function P upon Q and in step 6 write the solution set Now let us see one example here we have to solve this rational inequality Now let us start with its solution Now to solve this rational inequality we will follow the steps which we have discussed earlier Now in step 1 we have to write the given inequality in standard form but here you can see right inside of the given inequality is 0 thus it is already in standard form Now let us start with step 2 Now the given inequality is in the standard form that is P upon Q is greater than equal to 0 Now in second step we have to find all real zeros of polynomials P and Q that is we have to find all real zeros of numerator and denominator of this fraction Now here polynomial P is equal to x square plus 4x plus 3 and polynomial Q is equal to x square minus 16 Now to find zeros of P and Q we will equate them to 0 and factorize So firstly let us equate P is equal to 0 that is x square plus 4x plus 3 is equal to 0 Now let us factorize it by splitting the middle term so this is x square 3x plus x plus 3 is equal to 0 which implies now from these two terms so it will be x into x plus 3 the whole and from these two terms 1 is common so it will be plus 1 into x plus 3 the whole is equal to 0 which further implies x plus 3 the whole into x plus 1 the whole is equal to 0 So we obtain x is equal to minus 3 or x is equal to minus 1 Now let us put x square minus 16 is equal to 0 So x square minus 16 is equal to 0 implies x square minus 4 square is equal to 0 which further gives x plus 4 the whole into x minus 4 the whole is equal to 0 that is by applying the formula of a square minus b square which is equal to a plus b the whole into a minus b the whole So we have x is equal to minus 4 or x is equal to 4 Thus the critical values are x is equal to minus 4 4 minus 3 and minus 1 So in the second step we have obtained the critical values Now in the third step we will plot these critical values on the number line Now this is the point minus 4 this is the point minus 3 this is the point minus 1 and this is the point 4 Now where you can see that we have used those dots to represent the points minus 3 and minus 1 But open dots to represent the points minus 4 and 4 because minus 4 and 4 are the denominator So they are restrictions and will not be included in solution set So there these open dots represent that these points are not included in the solution set These closed dots represent that the points minus 3 and minus 1 are included in the solution set Now here you can see these 4 points divide the number line into 5 parts So we have 5 intervals First is open interval minus infinity to minus 4 that is this yellow shaded portion of the number line Then this pink shaded portion of the number line gives us the second interval Which is open interval minus 4 to minus 3 Then next interval is open interval minus 3 to minus 1 Then open interval is the open interval minus 1 to 4 The first interval is the open interval 4 to infinity Now in step 4 we will choose first number in each interval of the function p Now let us take the first interval that is the open interval minus infinity to minus 4 Now we can write it as x is less than minus 4 Now let us choose any value lying in this interval So here let us take x is equal to minus 5 that lies in this interval Now let us put x is equal to minus 5 in the given inequality So we have minus 5 whole square plus 4 into minus 5 plus 3 whole upon minus 5 whole square minus 16 is greater than equal to 0 Which implies 25 minus 20 plus 3 whole upon 25 minus 16 is greater than equal to 0 And this implies 8 upon 9 is greater than equal to 0 Which is true So this interval is the solution set of the given inequality Now let us take the second interval that is the open interval minus 4 to minus 3 Or we can write it as minus 4 is less than x is less than minus 3 Now let us take any value lying in this interval So here let us take x is equal to minus 3.5 Now let us put x is equal to minus 3.5 in the given inequality So we have minus 3.5 whole square plus 4 into minus 3.5 plus 3 4 upon minus 3.5 whole square minus 16 is greater than equal to 0 Now on simplifying we get 1.25 upon minus 3.75 is greater than equal to 0 Which is false So this interval is not the solution set of the given inequality Now let us take the next interval that is the open interval minus 3 to minus 1 Or we can write it as minus 3 is less than x is less than minus 1 Now here let us take any value lying in this interval So let us take x is equal to minus 2 Now again let us put x is equal to minus 2 in the given inequality Now when we put x is equal to minus 2 in the given inequality We get 1 upon 12 is greater than equal to 0 Which is true So this interval is the solution set of the given inequality Now let us take the next interval that is the open interval minus 1 to 4 Or we can write it as minus 1 is less than x is less than 4 Now here let us take x is equal to 0 Now when we put x is equal to 0 in the given inequality We get 3 upon minus 16 is greater than equal to 0 Which is false So this interval is not the solution set of the given inequality Now let us take the next interval that is the open interval 4 to infinity Or we can write it as x is greater than 4 Now let us take every value that lies in this interval So here let us take x is equal to 5 Now put x is equal to 5 in the given inequality Now when we put x is equal to 5 in the given inequality We get 48 upon 9 is greater than equal to 0 Which is true So this interval is the solution set of the given inequality Now in the next step we can construct sin chart Now on number 9 we see that in the interval that is the open interval minus infinity to minus 4 We have positive sign for the function e upon q Then in the open interval minus 4 to minus 3 we have negative sign for the function e upon q Now in the open interval minus 3 to minus 1 We have positive sign for the function p upon q Then in the open interval minus 1 to 4 We have negative sign for the function p upon q And in the open interval 4 to infinity We have positive sign for the function p upon q Now in step 6 we have to write the solution set Now here you can see that the function p upon q is positive In the open intervals minus infinity to minus 4 Minus 3 to minus 1 and 4 to infinity So we have these three intervals in the solution set And also we know that the points minus 3 and minus 1 are included in the solution set Because the given inequality is greater than or equal to And also these are That is minus 3 and minus 1 are zeros of numerator So solution set is Open interval minus infinity to minus 4 union Close interval minus 3 to minus 1 union Open interval 4 to infinity Now let us discuss exponential inequality Now we will see how can we solve an exponential inequality graphically Now suppose we have to solve 7 raised to power x is less than 12 for x Now here you can see that the right hand side cannot be written As with same ways 7 Thus we use graph to solve it For using graph calculator we see that this is the graph of the equation y is equal to 7 raised to power x Now here you can see that y is equal to 12 Then x is equal to 1.277 Then x is greater than 1.277 Then y is equal to 7 raised to power x Is also greater than and when x is less than 1.277 Then y is equal to 7 raised to power x Is also less than 12 Thus from graph we have 7 raised to power x is less than 12 x is less than 1.277 Thus solution set is The open interval minus infinity to 1.277 Now suppose we have to solve 7 raised to power x is less than 343 Now here you can see that right hand side Can be written in parts of base 7 Thus we have 7 raised to power x is less than 7 raised to power 3 Now we can equate the parts Because we have same ways Thus x is less than 3 So for x is less than 3 We have 7 raised to power x is less than 343 So its solution set is The open interval minus infinity to 3 So in this session we have discussed Rational and exponential inequalities And this completes our session Hope you all have enjoyed the session