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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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You need to be careful of division by 0. If $f(a)\ne0$, this is correct. –  David Mitra Jan 18 '12 at 14:41
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More precisely this is true for local maxima and minima, because $x\mapsto1/x$ is continuous and decreasing everywhere it is defined. However the function $x\mapsto\sin x$, defined whenever $\sin x\neq 0$, has many global maxima and minima, while its reciprocal has none. –  Marc van Leeuwen Jan 18 '12 at 14:50

3 Answers 3

up vote 3 down vote accepted

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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+1 I think this is the only correct answer. All the other answers assume differentiability, which might be incorrect (think of a V-shaped function). –  aelguindy Jan 18 '12 at 14:55
    
+1 Correctness is indeed a good thing. In addition, it is nice to maintain contact with the ground. Symbol manipulation tends to separate us from the eometry. –  André Nicolas Jan 18 '12 at 16:11
    
I need to use this result in a Physics report I am writing at University. Do you know any books or papers which I could reference which state this result? –  Josh Jan 19 '12 at 17:00
    
@Josh I doubt it's stated anywhere as a theorem (I do not know of any), it's more of an "exercise". But, the above argument is elementary enough, I think, to just be given in the report. –  David Mitra Jan 19 '12 at 17:15

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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