 Hello and welcome to the session. In this session, we are going to discuss the following question and the question says that show that the function f of x is equal to x cube minus 3 x square plus 3x minus 5 has a real 0 between 2 and 3. Use the intermediate value theorem to find an approximation for this 0 to the nearest tenth. Before starting the solution of this question, we should know our result and that is intermediate value theorem. According to this theorem, if f of x is a polynomial function and f of a and f of b have different signs, then there is at least one value c between a and b such that f of c is equal to 0. With this key idea, we shall move on to the solution. In this question, we are given a polynomial function f of x which is equal to x cube minus 3x square plus 3x minus 5 and we have to show that it has a real 0 between 2 and 3. Now let us find the value of the given function f of x at x is equal to 2 and x is equal to 3. So, f of 2 will be equal to 2 cube minus 3 into 2 square plus 3 into 2 minus 5. This is equal to 8 minus 3 into 4 plus 3 into 2 minus 6 minus 5 which is equal to 8 minus 12 plus 6 minus 5 and this is equal to minus 3. Similarly, f of 3 will be given by 3 cube minus 3 into 3 square plus 3 into 3 minus 5 and this is equal to 27 minus 3 into 9 plus 3 into 3 that is 9 minus 5. So, this is equal to 27 minus 3 into 9 that is 27 plus 9 minus 5. On solving it further, we get 4. Here we see that f of 2 is equal to minus 3 and f of 3 is equal to 4 and there is a sign change between f of 2 and f of 3. Then according to the intermediate value theorem given in the key idea, there is at least one value between 2 and 3 that is a zero of this polynomial function. Now, we check functional values at intervals of 1 tends for a sign change. For this, we first find the value of f of 2.1 and this is given by minus 2.669 that is here we have put the value of x as 2.1 in this function. Similarly, we find f of 2.2 and it is given as minus of 2.272. Now, f of 2.3 is given by minus 1.803, f of 2.4 is equal to minus 1.256, f of 2.5 is equal to minus 0.625, f of 2.6 is given by 0.096. Again here we have a sign change. So, now it will be x is equal to 2.5 or x is equal to 2.6. Now, we want to find the zero to the nearest tenth and we cannot necessarily go by which functional value is closer to zero. Further, let us subdivide this interval from 2.5 to 2.6 and calculate the functional values at intervals of one hundredth for a sign change. So, now we have f of 2.51 and its value is minus 0.557049. Similarly, f of 2.52 is equal to minus 0.488192, f of 2.53 is equal to minus 0.418423, f of 2.54 is equal to minus 0.347736, f of 2.5 is equal to minus 0.418423, f of 2.54 is equal to minus 0.418423, f of 2.54 is equal to minus 0.347736, f of 2.55 is equal to minus 0.27645, f of 2.56 is equal to minus 0.203584, f of 2.57 is equal to minus 0.130107, f of 2.58 is equal to minus 0.003584, f of 2.57 is equal to minus 0.130107, f of 2.58 is equal to minus 0.055688, f of 2.59 is equal to 0.00253. Now here again, we have a sign trained between successive hundreds. This means there is a 0 in between 2.58 and 2.59, since it would be slightly above 2.58 so they required 0 to nearest tenth would be 2.6. This is the required answer. This completes our session. Hope you enjoyed this session.