MHB Prove Monotony of Function: $f$ Strictly Decreasing

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A function is defined as strictly decreasing if for any two points a and b, where b > a, it holds that f(b) < f(a). Given that f is differentiable on the interval from (-∞, 0) to (0, ∞) and f'(x) < 0 for all x except at 0, where f'(0) = 0, it can be shown that f must be strictly decreasing. By assuming f is not strictly decreasing, one can find points a and b such that f(b) ≥ f(a), leading to a contradiction when applying the mean value theorem. This theorem implies that the average rate of change between a and b must be non-negative, contradicting the condition that f'(x) is negative. Therefore, the function f is proven to be strictly decreasing.
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Let $f$ be differentiable from $(-\inf,0)$ to $(0,\inf)$ and let $f'(x)<0$ for all real numbers except 0 and $f'(0)=0$. Prove that f is strictly decreasing.
 
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You might want to begin by stating the definition of a decreasing function. Then consider some examples.
 
I would use "proof by contradiction". Suppose f is NOT strictly decreasing. Then there exist a, b, b> a, such that f(b)\ge f(a). So $f(b)- f(a)\ge 0$. Since b> a, b- a> 0 so $\frac{f(b)- f(a)}{b- a}\ge 0$. Now use the "mean value" property.
 

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