Frustrated with Functions

[jinx24]

I am having a tough time with my homework assignment and I'm hoping someone can steer me in the right direction. Thank you so much.

f(x)=4-3x Find the following:

f(1/x)

1/f(x)

f[f(x)] I really don't know where to start on this one

x^2f(x)

f(-x) Is this the same as f(-1)?

-f(x)

-f(-x)

I know, or atleast I think I do, you just plug it into the x in 4-3x. I don't know why I am so confused. Thank you again for helping me.

 

[Gene]

You've sorta got it. You replace the x in 4 - 3x with whatever ? is in f(?). f(!)=4-3!
f(1/x) = 4-3(1/x)

f(-x) = 4-3(-x) = 4+3x
Try the rest and show your work.
-------------------
Gene

 

[jinx24]

Thank you so much for pointing me in the right direction. I guess I am trying to make this harder than it really is. I am still confused with the ones that have a number in front of the f(x) like 2f(x).

I came up with:
Given: f(x)= 4-3x

2f(x)=
= 2(4-3x)
= 8-6x

Also, f(1/x) and x^2f(x) and f[f(x)] are giving me trouble.
f(1/x)= 4-3(1/x)

x^2f(x)= x^2(4-3x) = 4x^2 - 3x^3 = x^-1 = 1/x

f[f(x)] = 4-3f(x)

I don't know if I should (or could) take these further...can they be simplified more? I know I can't solve for x because there is no = sign

Thanks again!

 

[Gene]

x^2f(x)= x^2(4-3x) = 4x^2 - 3x^3 = x^-1 = 1/x
Your first step is good, but
4x^2 - 3x^3 = x^-1 = 1/x
taint true. Suppose x=0. You are saying 0 = 1/0

f[f(x)]
Just follow the rule but do it starting with the inner most term.
f[f(x)] = f[4-3x] = 4-3(4-3x)
Your answer would come out the same but the path was shakey. Another rule (PEMDAS) says do inner ()s first.