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.

Questions:

1. Find the inverse of each of the following:

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

Solution:

i) Switch the y and the x variables.

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

ii) Solve for y.

3[x=(y+3)/3]3
3x= y+3
3x-3= y
f-1(x)= 3x-3

b) g(x)= ³√x-1

Solution:

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
Solution:

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?

Solution:

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".

Solution:

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)*(3-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?

Solution:

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.

Solution:

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)

3 comments:

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.