If $a

Question

I have the following GRE question that I have some trouble seeing.

If $g$ is a function defined o the open interval $(a,b)$ such that $a < g(x) < x$ for all $x \in (a,b)$, then $g$ is

A) an unbounded function

B) a nonconstant function

C) a nonnegative function

D) a strictly increasing function

E) a polynomial function of degee 1

I answered that D), because I thought I could take the derivative on the inequality $a < g(x) < x$ and get $0< g'(x)<1$, showing that the equation is strictly increasing. However the answer says it should be B) and I don't really see how they concluded this. Could anyone help me with this problem? Thanks in advanced!

Umm.... where did you get the idea that if $f(x) < g(x)$ that $f'(x) < g'(x)$ that is simply not true and not justifiable. — fleablood, Oct 23 '16 at 00:43
You've made a serious blunder: $f(x)<g(x)$ does not imply $f'(x)<g'(x)$. — symplectomorphic, Oct 23 '16 at 00:44
For example $x<3-x$ on $(0,1)$ but $1\nleq-1$. In essence, any constant on the first inequality has no effect on the second — Emilio, Oct 23 '16 at 00:52
@Emilio: even simpler, $1<2$ but as functions their derivatives are both identically zero. — symplectomorphic, Oct 23 '16 at 00:55

fleablood · Accepted Answer · 2016-10-23T01:13:07.247

1

Suppose $g(x) = c $ is constant. Than $a < c < x$ for all $x\in (a,b)$. But there must be some $y \in (a,c)$. $a < y < c=f(y)$. A contradiction.

So $g$ is not constant

====

It's worth noting for $g(x) \not \in (a,b)$ we can have $g(x)$ do anything. So C,D, E aren't possible answers. A) and B) are only possible if the are "forced" to be unbounded or non-constant on $(a,b)$. For A) $g(x)$ is actually forced to be bounded so A) is not correct.

It's also worth noting $g(x)$ need not be increasing on $(a,b)$ it can bounce around all it wants in $a < f(x) < x$.

edited Oct 23 '16 at 01:13

answered Oct 23 '16 at 00:46

fleablood

130,341

sorry for my post. I was typing on my silly phone, while yours went up. – RJM Oct 23 '16 at 00:52
Great minds and all that... – fleablood Oct 23 '16 at 01:07
You have a typo, $c=f(y)$ instead of $g(y)$ – Emilio Oct 23 '16 at 18:34
How can $g(x)$ not be in $(a,b)$? The function is only defined for $x\in(a,b)$, and for all such $x$, $a\lt g(x)\lt x\lt b$. – Barry Cipra Oct 27 '16 at 18:18
Hmm, the first three or four times I did not read this as $g(x)$ was only defined (a,b) but read it as a real function defined everywhere but with a certain behavior on (a,b). Not sure why I read it as such. Something about the wording made me feel it wasn't that sophisticated/abstract. But you are right. of g:(a,b)->R then g exists nowhere else. If it only exists on (a,b) then A) is false as it is bounded, C) could be false as a or b could be negative, E has no reason to be true. D) the function can fluctuate in its bound. – fleablood Oct 27 '16 at 18:26

score 0 · Answer 2 · answered Oct 23 '16 at 00:49

0

If $x \gt g(x)$ = $c \gt a$, where $c$ is constant, than $\exists y$ $\in$ (a,b), such that $y \lt c$. Therefore, $g(y) $ = $c \gt y$.

answered Oct 23 '16 at 00:49

RJM

1,127

score 0 · Answer 3 · edited Apr 13 '17 at 12:21

0

Differentiation is not monotonic, unlike integration.

GRE Subject Test in Mathematics - Where can I find related past papers, solutions to those, sample tests, advice, books, apps or other resources?

edited Apr 13 '17 at 12:21

Community

1

answered Oct 23 '16 at 03:56

BCLC

14,197

If $a

3 Answers3