# Homework Help: Increasing / decreasing functions

1. Nov 14, 2012

### Darth Frodo

I juat have a quick question.

Which of these is correct;

f(x) is one-to-one $\Rightarrow$ f'(x) is increasing

f(x) is one-to-one $\Rightarrow$ f'(x) is non-decreasing.

I think number one but I'm just not 100% sure.

Thanks

2. Nov 14, 2012

### Vargo

What about f(x) = arctan (x).

3. Nov 14, 2012

### Ray Vickson

It the function f1(x) = 1 - exp(-x) 1:1? If f1'(x) increasing? Is the function f2(x) = exp(x) 1:1? is f2'(x) increasing? Is f3(x) = arctan(x) 1:1? Is f3'(x) increasing everywhere? Is it decreasing everywhere?

RGV

4. Nov 14, 2012

### Darth Frodo

arctan in 1:1 and it looks as if it's always increasing.

What I want to know is that if a function is non-decreasing (i.e. a straight line) is it 1:1

5. Nov 14, 2012

### Zondrina

Suppose f(x) is a one to one function. So if f(x) = f(y) then x = y.

RGV gave a good example of a function that shoots down your theory, either one of them since they both mean the same thing.

Take f(x) = 1 - e-x. f'(x) = e-x.

So first note that f(x) is 1-1 and increasing. Now, f'(x) on the other hand is actually decreasing.

So it is not true that : If f(x) is 1-1, then f'(x) is decreasing.

To answer your other question about "what if f is a straight line" consider f(x) = c for some constant real number c.

There are infinitely many choices of x that give the same value for f, so the function is not 1-1.

6. Nov 14, 2012

### Darth Frodo

Ah right thanks Zondrina.

Yeah apologies for the confusion but I wrote it down incorrectly. I meant always increasing or always decreasing.

Thanks for the clarification

7. Nov 14, 2012

### Vargo

I don't mean to belabor the point, but if your function is 1-1 it need not be increasing or decreasing. e.g. Let f(x) = 1/x for negative x, f(x) = x for x greater than or equal to 0.

However, if f(x) is continuous and 1-1, then it is true that f must either be increasing or decreasing.

Also, if f is differentiable, and 1-1, then either $f'(x) \geq 0$ for all x. OR $f'(x)\leq 0$ for all x.

8. Nov 14, 2012

### Darth Frodo

Yeah, I was referring to Continuous functions, apologies.

Regarding your last point, if f'(x) = 0 doesn't this mean that f is not 1-1?

9. Nov 14, 2012

### Vargo

It depends on the set of values x such that f'(x)=0. The derivative can be zero at isolated points. For example, if f(x)=x^3, then f'(0)=0. You could even come up with examples where f'(x)=0 for an infinite number of x values. For example, if f(x)=x-sin(x), then f'(x)=1-cos(x) which equals 0 at all integer multiples of 2*pi, but is otherwise positive.

It is even possible for f'(x) to be zero at an infinite number of points in a finite interval. For example, suppose
$f'(x) = x^2 sin^2(\pi / x)$
This is 0 at all points x=1/n, but otherwise positive.

The technical requirement is that if f' ≥ 0 then f is strictly increasing if and only if the set of points x such that f'(x)=0 has measure 0.

Last edited: Nov 14, 2012