# If a function $f$ is decreasing on its domain then would its inverse be increasing or decreasing?

I have a question concerned the inverse of a function $f$ and the sign of its derivative.

If we are given a function $f$ that is decreasing on its domain, would its inverse $f^{-1}$ be increasing or decreasing.

I thought that this would be decreasing:

Because the formula for the derivative of the inverse is: $$\left(f^{-1}(x)\right)' = \frac{1}{f'\left(f^{-1}(x)\right)}$$

Since $f$ is decreasing, $f'< 0$, so $\left(f^{-1}(x)\right)' < 0$, so the inverse $f^{-1}$ would be decreasing.

However, I find that I am incorrect. why?

EDIT

This is the exact question

True or false: If $f$ is decreasing on its domain, then $f^{-1}$ is decreasing on its domain

The answer was false and it put "$f^{-1}$ would be increasing."

• maybe the fact it is not specified if the function is striclty decreasing? – Bento Aug 9 '14 at 12:06
• I think you are correct. A decreasing invertible function will have a decreasing inverse. You can actually make a more general argument that doesn't rely on differentiability of $f$ -- as $f$ may not in fact be differentiable. – paw88789 Aug 9 '14 at 12:07
• You are correct - the inverse of a decreasing function is also decreasing. – Amitai Yuval Aug 9 '14 at 12:07
• if it is not striclty decreasing, the inverse obviously does not exist. – Bento Aug 9 '14 at 12:14
• You didn't supposed that $f$ is derivable. But if you suppose $f$ derivable, and $f'(x)\neq 0$ for all $x$, then your proof is correct (but these are very strong hypothesis...) – idm Aug 9 '14 at 12:18

Yes, decreasing ! Indeed, let take $x\leq y$. By surjectivity, $x=f(u)$ and $y=f(v)$ for a certain $u$ and a certain $v$. Then, $f(u)\leq f(v)$ by hypothesis and so $u\geq v$ because $f$ is decreasing. By bijectivity, $u=f^{-1}(x)$ and $v=f^{-1}(y)$, therefore $f^{-1}(x)\geq f^{-1}(y)$.

Q.E.D.

• I added the actual question look above – Varun Iyer Aug 9 '14 at 12:26
• The issue has little to do with derivatives—your answer here is just what I would have said. – Lubin Aug 9 '14 at 12:53
• My precedent proof was wrong (I shown that $$\forall x,y: x\leq y, \exists u,v: u\leq v\implies f^{-1}(u)\geq f^{-1}(v)),$$ here is the correct proof. – idm Aug 9 '14 at 16:04

Can you explain why you think that you are incorrect? I also think that this would be decreasing. For $x>y$, we have $f(x)<f(y)$. Suppose that $f^{-1}(x)\geq f^{-1}(y)$. Since $f$ is decreasing, we have $f(f^{-1}(x))\leq f(f^{-1}(y))$, i.e., $x\leq y$. It is a contradiction.

• I added the actual question look above – Varun Iyer Aug 9 '14 at 12:25

Let it be that $A,B$ are subsets of $\mathbb{R}$ and that $f:A\rightarrow B$ and $g:B\rightarrow A$ are inverses of eachother.

Then $f\circ g=1_{B}$ and $g\circ f=1_{A}$.

Let it be that $f$ is decreasing on $A$.

If $b,b'\in B$ with $b<b'$ then $f\left(g\left(b\right)\right)=b<b'=f\left(g\left(b'\right)\right)$ implying that $g\left(b\right)>g\left(b'\right)$.

This justifies the conclusion the $g$ is decreasing on $B$.

If the question is the one specified later, then $f^{-1}$ may not exist for the reason that I said. Thus "There exists an $f^{-1}$ inverse of $f$ and $f^{-1}$ is increasing or decreasing" is a false statement, under the assumptions made. So it is not wrong to say that it is false, under the assumptions made, while it is wrong to say that $f^{-1}$ is increasing (assuming the existence of such an inverse).

It is false to say that there exists always an $f^{-1}$, under the assumptions made. Decreasing implies {strictly decreasing or not}: If the first case is true $f^{-1}$ can not exist, and we have cases in which the first case is true. If the question had been specified assuming the existence of the inverse as hypothesis, then automatically the implications of assuming decreasing $f$ would have been restricted to the striclty decreasing case. But this assumption was not set (and is not assured because of the decreasing one).

(But probably that existence was implicitly assumed... in this case the answer would be obviously "true" without needing differential calculus.)