September 21, 2006

Scribe Post: Day 10

Hi guys! This is Jann and I'll be your scribe for today.
As usual, we started our class with 3 problems.


1. Find the inverse of each of the following:

a) f(x)= (x+3)/3


i) Switch the y and the x variables.

y= (x+3)/3
x= (y+3)/3

ii) Solve for y.

3x= y+3
3x-3= y
f-1(x)= 3x-3

b) g(x)= ³√x-1


y= ³√(x-1)
(x)³= [³√(y-1)]³
x³= y-1
x³+1= y
g-1(x)= x³+1

c) h(x)=1/x

y= 1/x
y(x= 1/y)y
xy= 1
y= 1/x

Note: y=x is called an identity function. It means you get back what you started with. Since the inverse of this function is the same as it's original, it's considered to be a special function.

Note: If you feed the inverse of a function to itself, you'll always get "x" as an answer. [f(f-1(x))=x]

2. Given: f(x)= 3x/(x+5)

a) How do you know that "f" has an inverse?


The easiest way to know if the function has an inverse is to simply graph it. The graph should look something like this:If this passes the horizontal line test, it means it has an inverse. Technically, it passes the line test.

b) Find the inverse of "f".


Note: It was mentioned earlier that if an inverse of a function is fed to its original, the answer will always be "x".

f(f-1(x))= x
f-1(x){3f-1(x)/[f-1(x) +5]= x}f-1(x)
3f-1(x)= xf-1(x) +5x
3f-1(x) - xf-1(x)= 5x
f-1(x)= 5x/(3-x)

Note: We substituted "f-1(x)" as the unknown variable. After solving for "f-1(x)", we ended up with it's inverse: f-1(x)= 5x/(3-x).

3. f(x)= x^3 + 0.2x

a) Is "f" invertible?


Note: Like in #2, we should graph the function to know if it is invertible. The graph should look like this:

The graph of this function is invertible because it passed the horizontal line test.

b) Sketch a graph with Domain [-1.5, 1.5] and Range [-1,1].

Note: To sketch this graph, we must adjust the window setting of the screen. To change the setting...

i) Press [WINDOW]
ii) Set "Xmin" to -1.5 and "Xmax" to 1.5
iii) Set "Ymin" to -1 and "Ymax" to 1.
iv) Finally, press [GRAPH].

You should have something like this:

c) Find f-1(0.45) to 2 decimal places.


Note: To find the value of "f-1(x)", we need to trace it in the calculator. But first, we need to change our calculator setting to "Parametric". It describes what happens to x and y seperately.

i) Press [MODE]
ii) Arrow down to "Func" the arrow right to "Par". Press [ENTER]
iii) Press [Y=] then enter X1T= T.
iv) Enter Y1T= T^3 +0.2T
v) Press [GRAPH]

After entering the function, Press [WINDOW]. You'll see "Tmin","Tmax", "Tstep". Change "Tmin" to -1.5 and "Tmax" to 1.5

Note: "Tstep" is the calculation done by the calculator every pixel. f-1(0.45) is an "input", so we want the "output" of the function. If we trace the graph to x= 0.45, y should equal 0.18 to 2 decimal places.

Additional Problem: Find f-1(0.55)

Solution: Simply arrow right 2 times. You should see x=0.55, y=0.27

Lastly, if there is anything that's not explained properly, please give a comment.

I forgot to mention the method to draw the inverse of a function on the graph. Here are the steps:

i)Press [2nd] then [PRGM]
ii)Press [8] that says "DrawInv" then press [ENTER].
iii)The command will appear on the home screen. Press [VARS] then arrow right once then press [1].
iv) Press [Y1] (depends where you entered your function). Press [ENTER]

It should show you the inverse of the function you entered in [Y=].

I would also like to remind everyone that there "might" be a quiz tomorrow or on monday. Be ready.

Thats everything we talked about in class, I think. The next scribe is.... umm.... Christian. (^.^V)


christian said...

HI! Thanks for picking me... I mean it's not like I'm plotting to do something against you Jann right??? XP

Anyway.. I just noticed that you referred to "y = 1/x" as the identity function. I think it's
"y =x". That's as far as I've gotten. TIIIRED. See ya guys.

jann said...

Hi Christian,
Your welcome! I edited the error you mentioned. Thanks for commenting! (^_^V)

lindsay said...

good job on your post jann! you put in a lot of effort into it i'm sure. i know those graphs are ... time consuming. =). your post made me understand that lesson a bit more.

be happy.