Proving Differentiability of f given g'(x) < 0 $\forall$ x

mathwizarddud · 2008-06-28T07:38:50+00:00

Suppose the real valued g is defined on \mathbb{R} and g'(x) < 0 for every real x. Prove there's no differentiable f: R \rightarrow R such that f \circ f = g.

Thread starter mathwizarddud
Start date Jun 28, 2008
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Differentiability

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The discussion centers on proving that if a real-valued function g is defined on the real numbers and its derivative g'(x) is less than zero for all x, then there cannot exist a differentiable function f: R → R such that the composition f(f(x)) equals g(x). This conclusion is drawn from the properties of monotonic functions and the implications of g being strictly decreasing due to its negative derivative.

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Jun 28, 2008

mathwizarddud

Suppose the real valued [tex]g[/tex] is defined on [tex]\mathbb{R}[/tex] and [tex]g'(x) < 0[/tex] for every real [tex]x[/tex]. Prove there's no differentiable [tex]f: R \rightarrow R[/tex] such that [tex]f \circ f = g[/tex].