AP Calculus Summer Homework

[purplekid]

part of my AP Calculus summer homework includes 8 questions on functions:

If: f(x)={(3,5),(2,4),(1,7)} g(x)= (x-3)[sup:3td985rz]1/2[/sup:3td985rz] h(x)={(3,2),(4,3),(1,6)} k(x)=x[sup:3td985rz]2[/sup:3td985rz]+5
determine each of the following:

1. (f+h)(1)=
2. (k-g)(5)=
3. (f[sup:3td985rz]o[/sup:3td985rz]g)(3)=
4. (g[sup:3td985rz]o[/sup:3td985rz]k)(7)=
5. f[sup:3td985rz]-1[/sup:3td985rz](x)=
6. k[sup:3td985rz]-1[/sup:3td985rz](x)=
7. 1/f(x)=
8. (kg)(x)=

I'm familiar with how to solve these problems when the functions are equations, as in g(x) and k(x), but how do i do them for f(x) and h(x), since there is no x in the ordered pairs?

I have a few ideas about how it might be done, but I'm not sure, so here's what I've got:

for number one:
would (f+h)(1)={(3,5),(2,4),(1,7),(3,2),(4,3),(1,6)} make sense?
my logic being, that if f(x) and h(x) have no x in their solutions, then neither would f(1) or h(1)

it also occurred to me that these ordered pairs might be solutions to an equation that I am supposed to find, but I don't know how to find it.

for number two:
(k-g)(5)= 30 - 2[sup:3td985rz]1/2[/sup:3td985rz]
I'm pretty sure that one's correct, because these are the problems i have familiarity with (f(x)=equation, type problems)

for number three:
this one really confused me, would it possibly be simply the ordered pairs for f(x)?

number four is another one that I understand

number five:
taking the inverse of the ordered pairs is really confusing me. I dont really have any idea how to do this one.

number seven is also confusing, but i understand 6 and 8

if someone could just explain how the ordered pairs play into this, it would really help. I understand the basics of functions, composite functions, etc., but the f(x) and h(x) are really throwing me for a loop.

Sorry this is so long, I just wanted to make it as clear as possible what I do and dont understand.
Thanks in advance!

 

[mmm4444bot]

Hi Purple Kid:

When we compose functions, I think it's important to be aware of each domain involved.

Domain of f: {1, 2, 3}

Domain of g: x >= 3

Domain of h: {1, 3, 4}

Domain of k: all Real numbers

In other words, functions f and h are defined only for three values of x each.

(f + h)(1) is a CONSTANT, not a variable, so there is no need to write any x.

(f + h)(1) is FUNCTION NOTATION for the constant f(1) + h(1).

f(1) = 7 and h(1) = 6.

Therefore, (f + h)(1) = 13.

Likewise, (k - g)(5) represents a constant, not a variable.

(k - g)(5) is function notation for the number k(5) - g(5).

k(5) = 30 and g(5) = sqrt(2).

Therefore, (k - g)(5) = 30 - sqrt(2).

(fog)(3) DOES NOT EXIST because g(3) is not in the domain of f.

(fog)(3) means f( g(3) ).

g(3) = 0, but f( 0 ) is not defined because (again) 0 is not in the domain of f.

(gok)(7) means g( k(7) ).

k(7) = 54

g( 54 ) = sqrt(51)

Therefore, (gok)(7) = sqrt(51).

I think you might need better understanding of function notation (i.e., the symbolism indicating a variable versus the symbolism indicating a constant).

You wrote that you understand exercises (6) and (8), so I'm not sure why you posted them.

For the inverse of function f, we simply switch x <--> y. Perhaps, you need to review the concept of function inverses, too.

f^(-1)(x) = {(7, 1), (4, 2), (5, 3)}

1/f(x) means divide f(x) into 1.

f(1) is 7, so 1/f(1) = 1/7.

f(2) is 4, so 1/f(2) = 1/4.

f(3) is 5, so 1/f(3) = 1/5

Therefore, 1/f(x) = {(1, 1/7), (2, 1/4), (3, 1/5)}

If I wrote anything that you do not understand, please reply with specific questions.

Cheers,

~ Mark