 Hi and welcome to the session, I am Deepika here. Let's discuss a question. Show that the function given by fx is equal to e raised to power 2x is strictly increasing on r. Let f be a continuous close interval a, b and differentiable on open interval a, b. Then increasing interval a, b if f dash x is greater than 0 for each x belongs to open interval a, b. That is a function is strictly increasing on an open interval where its derivative is positive. So this is the key idea behind our question. So let's start the solution given x is equal to e raised to power 2x. Therefore, f dash x is equal to 2 into e raised to power 2x. Now we know that e raised to power x is equal to 1 plus x plus x square over 2 factorial plus x cube over 3 factorial plus so on. So therefore, f dash x is equal to 2 into 1 plus 2x plus 2x square upon 2 factorial plus so on. Now when x is greater than 0 dash x, this is also greater than 0. Hence function increases when x is greater than 0. When x is equal to 0, f dash x is equal to 2 which is also greater than 0. Hence function increases when x is equal to 0. x is less than 0, x is equal to minus y where y is a positive quantity. Therefore, f dash x is equal to e raised to power minus 2y and this is equal to 2 over e raised to power 2y and this is equal to 2 over some positive quantity. This implies f dash x is greater than 0 when x is less than 0 thus f dash x greater than 0 for all x belongs to r. Hence according to our key idea function, e raised to power 2x is strictly increasing on r. Hence proved, I hope the question is clear to you by and take care.