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Many-One into Function

Let f be defi...

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

f

be a twice differentiable function defined on

R

f (0) = 1,

f^{'} (0) = 2

f^{'} (x) \neq 0

x \in R .

∣ ∣ ∣ \begin{matrix} f (x) & f^{'} (x) f^{'} (x) & f^{''} (x) \end{matrix} ∣ ∣ ∣ = 0,

x \in R,

then the value of

f (1)

lies in the interval

f

be a twice differentiable function defined on

R

f (0) = 1,

f^{'} (0) = 2

f^{'} (x) \neq 0

x \in R .

∣ ∣ ∣ \begin{matrix} f (x) & f^{'} (x) f^{'} (x) & f^{''} (x) \end{matrix} ∣ ∣ ∣ = 0,

x \in R,

then the value of

f (1)

lies in the interval

f : (0, \infty) \to R

be a differentiable function such that

f^{'} (x) = 2 - \frac{f (x)}{x}

x \in (0, \infty)

f (1) \neq 1

f : R \to R

be twice continuously differentiable. Let

f (0) = f (1) = f^{'} (0) = 0

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