 Hello friends, welcome to the session I am Malika, we are going to discuss Lagrange's mean value theorem. Our given question is very quiet mean value theorem if fx equal to x square minus 4x minus 3 in the interval av where a equal to 1 and b equal to 4. Before starting with the solution I would like to tell you what is Lagrange's mean value theorem. Now Lagrange's mean value theorem if a function fx is number first continuous in the closed interval av second, derivable in the open interval av and third f a is not equal to fb then there exists at least one real number c being the element of open interval av such that f dash c equal to fb minus f a upon b minus a that is tangent at the point c on the graph of y equal to fx is parallel to the chord. Let's start with the solution we have given fx equal to x square minus 4x minus 3. Now we see that fx being a polynomial function is continuous in the closed interval 1 4 hence it is differentiable in the open interval 1 4. So now since fx is a polynomial function it is continuous closed interval 1 4 therefore it is differentiable in open interval 1 4. We will find f1 which is equal to 1 minus 4 minus 3 equal to minus 6 then f4 which is equal to 16 minus 16 minus 3 equal to minus 3 this implies f1 is not equal to f4. Now we see that all the three conclusions all good like branches mean value theorem. So this implies that there exists a point c which being the element of open interval 1 4 where dash c equal to fb minus f a upon b minus a. This can be written as f dash c equal to f4 minus f1 upon 4 minus 1. So let this be our course equation now we are given that equal to x square minus 4x minus 3. Now we will find f dash x which would be equal to 2x minus 4 therefore f dash c equal to 2c minus 4. Now putting this value in equation first we get 2c minus 4 equal to f4 is minus 3 minus and f1 is minus 6 upon b minus a that is 3. So this implies minus 4 equal to minus 3 plus 6 upon 3 which is equal to 1. This implies 2c equal to 5 this implies c equal to 5 by 2. Here we see that c equal to 5 by 2 being the element of open interval 1 4. If you understood the solution and enjoyed the session good bye and take care.