 Hello and welcome to the session. Let's work out the following question. It says, does there exist a function which is continuous everywhere, but not differentiable at exactly two points. Justify your answer. Let's now move on to the solution and that is the function fx given by mod x plus mod of x minus one and this can be again written as minus x plus minus of x minus one if x is less than zero and it is x plus minus of x minus one if x lies between zero and one plus x minus one if x is greater than one. This is where the definition of mod x which says it is minus x if x is less than zero and it is x if x is greater than zero similarly mod of x minus one will be minus of x minus one if x is less than one and it is x minus one if x is greater than one. One combining these two functions we get this for this x is less than zero but x is also less than one so this function is minus of x minus one here also x is less than one so this function is minus of x minus one but in this case both x is greater than one and of course x is greater than zero so the function is therefore fx is equal to minus x minus x is minus two x plus one so the function is one minus two x if x is less than zero and it is plus x minus x is zero plus one that is one if x lies between zero and one that is x plus x is two x minus one if x is greater than one. Now we will show that this function is continuous everywhere now we know that mod x is continuous everywhere of x minus one is continuous everywhere because we know that the mod function is continuous therefore their sum is continuous everywhere so this implies fx is continuous everywhere so that the function is not differentiable at exactly two points so in first show that the function is not differentiable at x is equal to zero we will check the differentiability at x is equal to zero so we will prove that the left hand limit is not equal to the right hand limit as h approaches to zero so h approaching to zero from the left hand side of f of zero plus h minus f of zero upon h now since h is approaching from the left hand side so f of zero plus h that is fh will be this limit h approaching from the left hand side that is h approaching to zero minus f of h that is one minus two h minus f of zero f of zero would be one minus two into zero that is one upon h and this would be equal to limit h approaching to zero minus one minus one is zero and this becomes minus two h upon h h gets cancelled and the limit would be minus two now we will check the right hand limit f of zero plus h minus f of zero upon h now since h approaching from the left hand side from the right hand side the function will be this limit h approaching to zero plus one minus f of zero f of zero would be one only one minus one upon h so this would be limit h approaching to zero plus of zero that is zero so limit h approaching to zero minus of f of zero plus h minus f of zero upon h is not equal to limit h approaching to zero plus f of zero plus h minus f of zero upon h so this implies function is not differentiable at now we will check the differentiability at x is equal to one so again we will check the left hand limit and the right hand limit f of one plus h minus f of one upon h now since h is approaching from the left hand side when x is less than one so the function will be this one that would be limit h approaching to zero minus one minus f of one will be one only so it is one minus one upon h so the limit would be now we find the right hand limit f of one plus h minus f of one upon h now since h is approaching from the right hand side when x is equal to one so the function will be this that is two x minus one so limit would be h approaching from the right hand side f of one plus h and the function is two x minus one so it would be two into one plus h minus one minus f of one that is two into one minus one upon h so this would be limit h approaching to zero plus of two plus two h minus one minus two minus one is one upon h h approaching to zero plus of two h upon h limit would be hand limit of f of one plus h minus f of one upon h is not equal to limit h approaching right hand side of f of one plus h minus f of one upon h so this implies fx is not differentiable x is equal to one hence the function is continuous everywhere but it is not differentiable at x is equal to zero and at x is equal to one so this completes the question on this session my for now take care have a good day