Let `f` be arry continuously differentiable function on `[a,b]` and twice differentiable on `(a.b)` such that `f(a)=f(a)=0 and f(b)=0.` Then

Let f be any continuously differentiable function on [a, b] and twice differentiable on (a, b) such that f(a)=f'(a)=0" and "f(b)=0 . Then-

Let f be any continuously differentiable function on [a,b] and twice differentiable on (a,b) such that f(a)=f'(a)=0 . Then

If f(x) is a twice differentiable function such that f(0)=f(1)=f(2)=0 . Then

If f(x) is a twice differentiable function such that f(0)=f(1)=f(2)=0 . Then

Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then :

Tet f be continuous on [a,b] and differentiable on (a,b). If f(a)=a and f(b)=b, then

Let f:Rto R be a twice continuously differentiable function such that f(0)=f(1)=f'(0)=0 . Then

To illustrate the Mean Value Theorem with a specific function, let’s consider f(x) = x^(3) – x, a= 0, b = 2 . since f is a polynomial, it is continuous and differentiable for x, so it is certainly continuous on [0, 2] and differentiable on (0, 2) such that f(2) – f(0) = f'(c)(2 – 0)

Let f (x) be twice differentiable function such that f'' (x) lt 0 in [0,2]. Then :

Let f: [a, b] -> R be a function, continuous on [a, b] and twice differentiable on (a, b) . If, f(a) = f(b) and f'(a) = f'(b) , then consider the equation f''(x) - lambda (f'(x))^2 = 0 . For any real lambda the equation hasatleast M roots where 3M + 1 is