## Odd Function Problem

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
burnbird16
Posts: 47
Joined: Sat May 22, 2010 1:29 am
Contact:

### Odd Function Problem

Yet another problem for your brilliant minds to ponder!

Simplify (f(x+h) - f(x))/h, where f(x) = 2x+3

Here's what I did, please tell me if I made an error.

2(x+h) +3 - 2x - 3 / h

2x + 2h + 3 - 2x - 3 /h

2h/h

2

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Odd Function Problem

Yet another problem for your brilliant minds to ponder!

Simplify (f(x+h) - f(x))/h, where f(x) = 2x+3

Here's what I did, please tell me if I made an error.

2(x+h) +3 - 2x - 3 / h

2x + 2h + 3 - 2x - 3 /h

2h/h

2

You really need to use parentheses or do this...

$\frac{2(x+h) +3 - (2x + 3)} { h}$

$\frac{2x+2h +3 - 2x - 3} { h}$

$\frac{2h} { h}=2$

burnbird16
Posts: 47
Joined: Sat May 22, 2010 1:29 am
Contact:

### Re: Odd Function Problem

I'm sorry, but I don't know how to do what you just did.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Odd Function Problem

I'm sorry, but I don't know how to do what you just did.

$\frac{2(x+h) +3 - (2x + 3)} { h}$

$\frac{2x+2h +3 - 2x - 3} { h}$

$\frac{2h} { h}=2$

Code: Select all

$$\frac{2(x+h) +3 - (2x + 3)} { h}$$ $$\frac{2x+2h +3 - 2x - 3} { h}$$ $$\frac{2h} { h}=2$$

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Odd Function Problem

you could have also did it this way...

[2(x+h) +3 - 2x - 3] / h

=[2x + 2h + 3 - 2x - 3]/h

=2h/h

=2

for h not equal to zero

burnbird16
Posts: 47
Joined: Sat May 22, 2010 1:29 am
Contact:

### Re: Odd Function Problem

Ok, thank you. So was I correct,then? If so, I have a few more I would like for you to look at, if you have time.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Odd Function Problem

Ok, thank you. So was I correct,then? If so, I have a few more I would like for you to look at, if you have time.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Odd Function Problem

... I have a few more I would like for you to look at, if you have time.
I'll be on for a little while...and I'm sure there are others here that can also help with this.

burnbird16
Posts: 47
Joined: Sat May 22, 2010 1:29 am
Contact:

### Re: Odd Function Problem

Of course, and I will do so, thank you very much ^^

However, I do have a few more problems that I need some help with.

I'll just list them off, as I don't even know how to start any of them.

30. a) The graph of a quadratic function has x-intercepts, -1 and 3, and a range consisting of all numbers less than or equal to 4. Determine an expression for the function. (I tried to reverse factor, but I got a positive parabola, not a negative one.)

31. Write as a single equation in x and y: x = t + 1, y = $t^2+t$

34. Express x in terms of the other variables in this picture: http://www.flickr.com/photos/46210274@N07/4918073428/ (I drew it myself, I know it's kind of small, but it is a replica of the drawings in the packet)

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Odd Function Problem

...
30. a) The graph of a quadratic function has x-intercepts, -1 and 3, and a range consisting of all numbers less than or equal to 4. Determine an expression for the function. (I tried to reverse factor, but I got a positive parabola, not a negative one.)
...

for this your function needs to look like...

$f(x)=k(x+1)(x-3)$

since you want the range to be less than or equal to 4 we know the parabola goes down so $k<0$

by symmetry the max will occur at x=1 (the middle of -1 and 3)

so plug 1 into the function and solve for k (ie solve $f(1)=4$)