#### Question

Let f defined on [0, 1] be twice differentiable such that | f"(x) | ≤ 1 for all x ∈ [0, 1]. If f(0) = f(1), then show that | f'(x) | < 1 for all x ∈ [ 0, 1] ?

#### Solution

If a function is continuous and differentiable and *f*(0) = *f*(1) in given domain *x* ∈ [0, 1],

then by Rolle's Theorem;*f*'(*x*) = 0 for some *x* ∈ [0, 1]

Given: |*f*"(*x*)| ≤ 1

On integrating both sides we get,

|*f*'(*x*)| ≤* x*

Now, within interval *x* ∈ [0, 1]

We get, |*f*' (*x*)| < 1.

Is there an error in this question or solution?

Solution Let F Defined on [0, 1] Be Twice Differentiable Such that | F"(X) | ≤ 1 for All X ∈ [0, 1]. If F(0) = F(1), Then Show that | F'(X) | < 1 for All X ∈ [ 0, 1] ? Concept: Increasing and Decreasing Functions.