For any continuous function $f(x)$ which has a maxima at $x = a$. Will the reciprocal of that function i.e. $g(x) \equiv 1/f(x)$ always have a minima at $x = a$ and if so can this be proven?
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No need to be fancy: Suppose $f$ has a local maximum value at $x=a$ and that $f(a)>0$. Then by continuity and the assumption that $x=a$ gives $f$ a local maximum value, there is a $\delta>0$ such that $f(x)>0$ and $f(x)\le f(a)$ for all $x\in I=(a-\delta,a+\delta)$. This implies ${1\over f(a)}\le {1\over f(x)}$ for all $x\in I$. Thus ${1\over f}$ has a local minimum value at $x=a$. I'll leave the case when $f(a)<0$ for you. |
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It will be a minimum. To prove this you have to compute the second derivative that is $$\frac{d^2g}{dx^2}=\frac{2[f'(x)]^2}{[f(x)]^3}-\frac{f''(x)}{[f(x)]^2}.$$ At the point $x=a$, being $f(a)\ne 0$, this just becomes $$\left.\frac{d^2g}{dx^2}\right|_{x=a}=-\frac{f''(a)}{[f(a)]^2}$$ and so this is negative if $f''(a)$ was positive and viceversa changing the nature of the extremum. |
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I think by inverse you mean reciprocal. I prove it local maxima for smooth functions. The claim is true, if you assume that $f$ is nowhere zero. Assume $f$ is smooth. Then de derivative of $g$ equals, using the chainrule $$g'(x)=-1/f(x)^2 f'(x)$$ And the second derivative is using also the product rule $$g''(x)=2/f(x)^3 f'(x)^2-1/f(x)^2 f''(x)$$. Assume that $c$ is a maximum of $f$, then $f'(c)=0$ and $f''(c)<0$. But then $g'(x)=0$ by the above formula, and $g''(c)=0-1/f(x)^2 f''(x)>0$. So $c$ is a minimum for your function $g$. |
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