Proving F is not an increasing function?

I'm not sure if I am understanding this question in my math book correctly. It says to let f be a function from R to R. Without using words of negation, write the meaning of "f is not an increasing function." Does this mean that I just need to rewrite the statement without using the word "not", or am I somehow supposed to prove that f is not an increasing function? If I am supposed to prove this, how would I go about that?

Thanks!

Re: Proving F is not an increasing function?

First, write the meaning, the definition, of f IS an increasing function. Then negate it, since it's NOT true that f is an increasing function.

Meaning of "f IS increasing" is given by "For all x1, x2 in R, with x1 < x2, have _______" (fill in some condition C(x1, x2, f)).

Meaning of NOT "f IS increasing" is therefore "There exists x1, x2 in R, with x1 < x2, such that _______" (now fill in the condition NOT C(x1, x2, f)).

Re: Proving F is not an increasing function?

Would it also be correct to say something along the lines of "If f is a function such that f'(x)<0, f must not be an increasing function"? I'm not sure when I'm allowed to use derivatives in proofs, and when I'm not.

Re: Proving F is not an increasing function?

You are asked to negate the property from the *definition* of an increasing function, but write this negation without the word "not."

Re: Proving F is not an increasing function?

Quote:

Originally Posted by

**mdlavey** Would it also be correct to say something along the lines of "If f is a function such that f'(x)<0, f must not be an increasing function"? I'm not sure when I'm allowed to use derivatives in proofs, and when I'm not.

Since a "general" function can't be assumed to have a derivative, you'll typically be told to assume that f is differentiable if the derivative of f is going to be used in whatever you're trying to show (like statements about function increasing). If you need f to be differentiable, but it isn't known before hand that it is (meaning it isn't part of the problem's "givens"), then you'll need to somehow demonstrate that it's differentiable prior to refering to its derivative.

That's all about math statements for a generic "function f". If you're given a specific definition for f, like f(x)=sin(2x-3)/e^x, then you're assumed to already know where it's differentiable (well, unless it's Calculus 1, in which case showing that it's differentiable might be the problem itself).

The bottom line is, for a generic "function f", you can't assume it's differentiable anywhere (or even that it's continuous, or bounded, or even that a one-sided limit exists anywhere - you simply can't assume anything about a generic function other than it is a function from its domain to it range.)

Re: Proving F is not an increasing function?

Quote:

Originally Posted by

**mdlavey** I'm not sure if I am understanding this question in my math book correctly. It says to let f be a function from R to R. Without using words of negation, write the meaning of "f is not an increasing function." !

It is impossible to answer this without know what the author means by *increasing.*

If it means $\displaystyle a < b \Rightarrow f(a) < f(b)$ then the answer is $\displaystyle \left( {\exists a} \right)\left( {\exists b} \right)\left[ {a < b \wedge f(a) \geqslant f(b)} \right]$

On the other hand, if it means $\displaystyle a \le b \Rightarrow f(a) \le f(b)$ then the answer is $\displaystyle \left( {\exists a} \right)\left( {\exists b} \right)\left[ {a \le b \wedge f(a) > f(b)} \right]$.