Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You need to be careful of division by 0. If $f(a)\ne0$, this is correct. – David Mitra Jan 18 '12 at 14:41
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
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.

share|cite|improve this answer
+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? – joshlk 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


At the point $x=a$, being $f(a)\ne 0$, this just becomes


and so this is negative if $f''(a)$ was positive and viceversa changing the nature of the extremum.

share|cite|improve this answer

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 $g$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.