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

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Suppose the real valued g is defined on \mathbb{R} and g&#039;(x) &lt; 0 for every real x. Prove there's no differentiable f: R \rightarrow R such that f \circ f = g.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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