# Increasing functions

1. Nov 24, 2008

### Firepanda

http://img135.imageshack.us/img135/9429/asdaauh6.jpg

I'm not too sure about this, I thought as a example function e^x, where I can say as x gets larger then so does e^x and then so does its derivative at those values of x.

But what if it was a function where as x increased the function was getting smaller, and so it would be a decreasing function.. or perhaps it doesn't matter and I should assume a function like that doesn't exist?

Is there any other fucntion with that property other than e^x?

Really confused about the second part, taking e^x as an example, then then inverse would be lnx, and its derivative 1/x. Then surely I'm supposed to assume all they are talking about here is e^x, since no other f'(x) can take the form of 1/x other than f(x) = lnx...

Last edited by a moderator: Apr 23, 2017 at 10:19 PM
2. Nov 24, 2008

### f(x)

since f(x)=f(x) and codomain is R+ f(x) is always > 0 hence increasing.

3. Nov 24, 2008

### Firepanda

Would that be considered a proof?

4. Nov 24, 2008

### HallsofIvy

Staff Emeritus
What f(x) said is not quite true. f is NOT into R+ and the f ' is NOT always "> 0" but close:
Your problem tells you that f(x) is never negative and that f'(x)= f(x) for all x. That is, f'(x) is never negative. What does that tell you?

5. Nov 24, 2008

### Firepanda

Ah I see, the second part though, so far I can say the slope of the inverse function would be given by 1/f'(x).

Can someone tell me if the question is asking for the derivative of the inverse function, or the inverse of the derivative?

Edit:

I see it's asking for the derivative of the inverse function, so what I did was this:

the slope of the inverse would be 1/f'(x) = dx/dy taking y = f(x)

then I went on to say that 1/f'(x) = (f -1)' (x) (derivative the inverse of f(x))

So howdo i show this is 1/x ?

Last edited: Nov 24, 2008
6. Nov 25, 2008

### Firepanda

gta hand this in in 30 mins, hoping someone will read this and help :(

7. Nov 25, 2008

### HallsofIvy

Staff Emeritus
You are confusing x and y in y= f(x) and x= f-1(y).

8. Nov 25, 2008

### f(x)

Hello sir,
Pardon me but could you plz explain why f(x) >= 0 and not f(x) > 0 since the codomain of f(x) has an interval (0,∞) exclusive while f(x) = f`(x)
Thanks

9. Nov 25, 2008

### HallsofIvy

Staff Emeritus
You are right, I was wrong. I misread $(0, \infty)$ as $[0, \infty)$.