Ellie's Blog for 3/4/13 & 3/5/13

Hey Guys! I'm blogging for Monday and Tuesday of this week.

On Monday, we started out by talking about function operations. These operations are addition, subtraction, multiplication, division and composition. Here's the notation for each of these operations:

Addition: f(x) + g(x) or (f+g)(x)

Subtraction: f(x) - g(x) or (f-g)(x)

Multiplication: f(x) * g(x) or (f*g)(x)

Division: f(x)/g(x) or (f/g)(x)

Composition: f(g(x)) or (f ◌ g)(x)

Next, we looked at these tables and evaluated some expressions:

For example, we can determine that:

f(3) = 5

g(4) = 3

(f-g)(4) = -11 because f(4)-g(4) = -8 - 3 which equals -11

(f/g)(2) = undefined because there is no f(2) in the table

Once we had a handle on that, we moved on to composition. 

First of all, what is composition?

Composing two functions means plugging one function into the other function or plugging the output of the "inside" function into the "outside" function. For example, if you're looking at f(x) composed with g(x) which is written f(g(x)) or (f ◌ g)(x), the "inside" function is g(x) and the "outside" function is f(x).

Now for some examples:

We looked at the functions f(x) = 6x+1 and g(x) = √x

In order to evaluate f(g(0)), first you look at the "inside" function which is g(0). When you plug 0 into the function g(0) =√x, the output is 0 so you plug this output in for g(0) so now you're evaluating f(0) which ends up equalling 1

Let's look at another example: 

f(g(25)) = 31 because g(25)=5 and f(5)=31

These are some specific examples, but to put this in more general terms:

f(g(x)) = f(√x) = 6√x+1

g(f(x)) = g(6x+1) =√6x+1

This means that if you're asked to find f(g(x)), you first plug in √x for g(x) and then you plug in √x for x in the other function as shown. The opposite is true for g(f(x)).

Then we moved on to doing some problems on the board:

This same substitution idea can be seen in the problem above

Here's another similar one:

Then we tried doing these problems in reverse order. Here's an example:

This problem was fairly straightforward because when you look at (x+1)³ you can see that (x+1) already looks like it's on the "inside" and can be plugged in for g(x) as shown above.

 And if g(x) = (x+1), then f(x) has to equal x³ in order for f(g(x)) to equal (x+1)³

Our homework was to do p. 294: 1,3,5,7,15,17

These problems are like what we did on the board and are good practice of these concepts.

On Tuesday, we talked about inverse relations. We started out by graphing 3x-7 and (x+7)/3 which are inverses of each other. The graph looked like this:

Then we talked about what makes two functions inverses.

- If two functions are inverses, then f(g(x)) = g(f(x)) = x

- If point (a,b) is on f(x), then (b,a) is on its inverse which is f ^-1(x)

We also discussed how f(x) and f ^-1(x) are symmetric over y=x, as illustrated in the graph below:

Then we had to determine the inverse of y = 1/2x - 5

In order to do this we had to first switch the x's and y's so then we had: x = 1/2y - 5

Then we solved for y and got that y = 2x + 10

Then we checked our work:

f(x) = 1/2x - 5

f ^-1(x) = 2x + 10

f(f ^-1(x)) = f(2x+10) = 1/2(2x+10)-5 = x

f ^-1(f(x)) = f ^-1(1/2x-5) = 2(1/2x-5)+10 = x

We know this is correct because if two functions are inverses, then f(g(x)) = g(f(x)) = x, as I said above. 

Alright, I think that covers what we've talked about for the past two days. If anyone has any questions or if I missed something, please let me know :)

And now for some more good math humor....