 Now it is minus 2. Is this differentiable at x equal to minus 1? Check the difference. What you may redefine the function? This will be in the graph. Yes, what is the conclusion? Absolutely correct, at 2 it is not differentiable but at minus 1 it is. You know why at minus 1, mod of it and it became continuous at here at minus 1. But it is not differentiable. This is just why I keep finding such functions. Whenever I am dealing with mod function, GIF function or any kind of a special function, always redefine it first. Redefining means getting rid of the mod and rid of those special symbols. Make it simple looking. Any operation should be done by making it simple looking. Now, this particular function. Science means very little, not sure. It is very simple. You just first write down the zeros of these factors on the left. So when x plus 1 becomes 0, at negative 1, second step is put 3 here and it will stay positive. That means any value you choose beyond 2, where it is 3, 4, 500, 500, 6, 7, 5, 100, whatever. The expression will always be positive. At this point. This is something which may be correct. If 2 is associated with a factor which has got an apple switch. So obviously a question will arise in your mind what if there was an even power on h minus 2? Are you getting this one? Now, so this will become negative. Now I am crossing this number minus 1. factor for which minus 1 is coming, x plus 1, this is called the baby curve science thing. When you try to know the sign of polynomial functions, in fact it can work for rational functions as well in some intervals of x. Now why am I doing this? Minus x minus 2 would remain x square minus x minus 2 if your x minus 2 was becoming negative, is it not? What will mod be to it? It will try to make it positive. How do you make the greatest sign in front of it? So you will get minus x minus 2 if your x square to belong to the interval minus 1 will do. Now most of you would ask, say you include it to here also, it does not matter because at 2 it is either plus 0 or minus 0 which is ultimately 0, 0 does not have a sign. Minus 1 also are included both the places, by the way this is not that there, this is minus 1. That is why at minus 1 I can include it at both the places because it will actually be 0. Now if I know this times this, so if I have x plus 1 here, can I say x plus 1 will mod it? Is it? Now what points will it test the differentiability and continuity? Which points will I test the differentiability? Now there is something called critical points, just like we had critical points in case of continuity. We do not sit and check for all the points in the world because there is something, we only check at those points where the function is changing its definition provided we know that the function is a safe function. For example this is a polynomial function and in the first property itself I stated that polynomial functions are continuous interval functions. But where it is changing the definition from one function to another function, there may be a break and hence there may be a non-differentiability or there may be a non-differentiability, getting this point done. So I will only check at those points which we refer to as the critical points. So x plus 1 mod x square minus x minus 2 equal to minus 1, the function was behaving as x plus 1 x square minus x minus 2. When it was between minus 1 and the x square minus x minus 2 and when my function was greater than equal to 2, it was again behaving as the previous function, this one. Is that right? Now tell me what are the critical points minus 1 and 2 because at these points my function is changing its definition. So let us check at minus 1, first thing first minus 1 will I be checking its continuity at minus 1? The main reason behind including minus 1 at both the places is because I think it is continuous. Polynomial and the problem is both are continuous, so they are product and directly jump to differentiability, but will I use my first principles? Oh my God, that would be too much of a work. So now welcome to the J word, welcome to the cognitive exam word. In school only you will do your first principle test. When you are actually solving it in CET or mid-style or J main, what are the exams we are preparing for, we will do a shortcut. What is a shortcut? If I have to find the left hand derivative at minus 1, just differentiate the function which is left of minus 1 and put minus 1. Which is the function left of minus 1? Is it this or is it this? The first one, differentiate it the normal way. Whatever rules you know, product, chain, quotient, whatever you know, differentiate the normal way, put x equal to minus 1 and tell me the answer that you get. So three, what do you plan to differentiate or you can use product rule? Of course, you give me the value, you give me the value, you differentiate this guy and put x equal to minus 1, give me the answer. Let's go over all this. I'm making a mistake. Half the glass, 2 x, x square minus 1, minus 2, it's becoming. Oh, I didn't get it wrong. Okay, answer. So what do you do with the point of view? Now, differentiate this guy, put x equal to minus 1 and tell me the answer. Before anything else. Zero, minus 1 because of the procedure. Please tell me, yes, let's confirm that, we know that's not different. First, let's confirm whether, yes, this guy, differentiate this at minus 9. Let it, let it, let others also confirm. Minus 9, minus 9, minus 9. I was assuming you were correct. What is the right hand derivative I do? Are they equal? No. No. And hence, we got differentiate. Is the ideal here? So, so much analysis is being done just for modding a function. Is that fine? Moving on, can I erase this? They just make sense? Yes. Differentiate, put 2. Differentiate this, put 2. You'll get left hand derivative. Differentiate this, put first principles here. It will become huge because of the powers involved. It works for smaller functions, which you normally would get in your school exam. But in complicated exams, do not use first principles at all. There's a negative. You're saying something, sir. I mean, they're essentially the same functions that the one in the middle has a negative. So, the slope of that would be negative. Yeah, of course. You're also finding it would be nice. Is that fine? Well, I'm not done with the properties yet. I think I did the third one, so far. Yeah, yeah, please, please.