Let f defined on -0- 1- be twice differentiable such that - f quot-x- - le- 1 for all x isin- -0- 1- If f-0- - f-1- then show that - f -39-x- - lt- 1 for all x isin- - 0- 1-

If a function is continuous and differentiable and f(0) = f(1) in given domain x isin; [0, 1],then by Rolle #39;s Theorem;f #39;(x) = 0 for some x isi ...

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Byju's Answer

Standard XII

Mathematics

Derivative of One Function w.r.t Another

Let f defined...

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].

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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.

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Q. 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].

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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].

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)| ≤ 1On integrating both sides we get,|f'(x)| ≤ xNow, within interval x ∈ [0, 1]We get, |f' (x)| < 1.

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.