Determine the value of this function

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The discussion revolves around solving two function-related problems. For part A, participants clarify that substituting x=2 leads to the equation 2f(2) - 3f(1/2) = 4, and suggest using x=1/2 to create a second equation for the unknowns f(2) and f(1/2). In part B, the recursive function f(n) = 2f(n-1) + 3f(n-2) is discussed, with advice to calculate intermediate values like f(3) and f(4) before finding f(5). One participant successfully solves for f(2) after applying the suggested method, indicating progress in understanding the concepts. Overall, the thread emphasizes the importance of recognizing relationships between function values to solve for unknowns.
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Homework Statement



A)If 2f(x) - 3f(1/x) = x^2 , determine f(2).

B)Let f be a function defined by f(n)=2f(n-1) + 3f(n-2), where f(1)=1 and f(2)=2. Determine f(5).

The attempt at a solution

I tried to plug in the 2 into all the x's but got the wrong answer. would really appreciate some help to get me understand this topic of functions. not my strong suite at all.
 
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It's best if you show your work [with steps]... even if it is the wrong answer.
 
ok.

for a) i am not sure how to do it, but here is my attempt.

as far as i understand, f(x) means just plug in a value for x.
f(2) i think is just x=2

2(2)-3(1/2)=2^2
4-3/2=4
8/2-3/2 doesn't equal 4.

so as you can see i am lost.

b)same thing for be, just put a value given for n.

f(5)=2f(5-1) + 3f(5-2)
f(5)=8+9
f(5)=17

that answer is also wrong.

some help would be greatly appreciated.
 
Moe_slow said:
A)If 2f(x) - 3f(1/x) = x^2 , determine f(2).

Moe_slow said:
ok.

for a) i am not sure how to do it, but here is my attempt.

as far as i understand, f(x) means just plug in a value for x.
f(2) i think is just x=2

2(2)-3(1/2)=2^2
4-3/2=4
8/2-3/2 doesn't equal 4.

so as you can see i am lost.

"f(x) means just plug in a value for x" ... is okay
"f(2) i think is just x=2"... ok... but this does not mean that necessarily f(2)=2

All you can say is
2f(x) - 3f(1/x) = x^2
with x=2 is
2f(2) - 3f(1/2) = 2^2
which has two unknowns f(2) and f(1/2).
However, with a clever choice a different value of x, you can get a second equation in the same two unknowns, which produces a system of two equations with two unknowns, which can be solved.
 
And for (b), calculate f(3) first. Then f(4) and f(5).
 
sorry, i tried to figure out what you two wanted me to do, but couldn't. i will try to find some other form of help online. thanks again
 
tried really hard to find some form of method that will help me understand but couldn't. these two question for some odd reason are placed at the end of the "introduction to functions" chapter of my precal textbook. they are not part of any homework or test or anything, i was just discombobulated with them and wanted some form of answer to help me learn. if someone could show me how they were solved would be great, they are not an assignment or test so i don't think it would violate any rules. just need some form of look into the steps to solve these. thanks.
 
Moe_slow said:
tried really hard to find some form of method that will help me understand but couldn't. these two question for some odd reason are placed at the end of the "introduction to functions" chapter of my precal textbook. they are not part of any homework or test or anything, i was just discombobulated with them and wanted some form of answer to help me learn. if someone could show me how they were solved would be great, they are not an assignment or test so i don't think it would violate any rules. just need some form of look into the steps to solve these. thanks.

did you understand my post?

Looking back at your first post:
"as far as i understand, f(x) means just plug in a value for x.
f(2) i think is just x=2"

f(x) is rule that says:
if you give me an x (say 2), I'll give you back a number based on what you gave me. It need not be the same number you gave me.

ex: f(x)=x+1
if you give me 1, f(1)=2
if you give me 2, f(2)=3
if you give me 3, f(3)=4

ex: f(x)=(x-2)^2
if you give me 1, f(1)=1
if you give me 2, f(2)=0
if you give me 3, f(3)=1

In the problem that you posed,
you don't know explicitly what f(x) is...
but you are given some hints based on equations that must be satisfied by this hidden function.
 
Last edited:
Moe, you pretty much show a lack of understanding of what a function is... f(3) =/= 3 unless f(x)=x, which you definitely don't know (and obviously isn't true, since you proved if it was it wouldn't work!)
 
  • #10
If you replace x with 1/x through out the equation, you get two equations with f(x) and f(1/x). Eliminate f(1/x) to get f(x) and solve your question.
 
  • #11
Moe_slow said:

Homework Statement



A)If 2f(x) - 3f(1/x) = x^2 , determine f(2).

B)Let f be a function defined by f(n)=2f(n-1) + 3f(n-2), where f(1)=1 and f(2)=2. Determine f(5).

The attempt at a solution

I tried to plug in the 2 into all the x's but got the wrong answer. would really appreciate some help to get me understand this topic of functions. not my strong suite at all.

It has already been pointed out to you that setting x= 2 gives 2f(2)- 3f(1/2)= 4. What do you get if you set x= 1/2? (That's the "clever choice" robphy was talking about.) Now you have two equations for the same two unknown values of f.

For b), do what phoenixthoth suggested: Since f(x)= 2f(n-1) + 3f(n-2), f(3)=2f(3-2)+ 3f(3-2)= 2f(1)+ 3f(2). What is that equal to? Now do the same thing with f(4) and f(5).
 
  • #12
i am not sure i understand the concept of the two unknowns. i tried to do it like linear systems but failed. i went through the textbook to try and find another question like these but there are none. does anyone have a link to a good guide that will help me understand the concepts?
 
  • #13
Thats exactly what it is! 2 equations with 2 unknowns, except your variables are f(2) and f(1/2). We have no clue what these can be, so you need to substitute or eliminate to solve for f(2). The hard part was noticing the trick of letting x=1/2. After that, its all algebra
 
  • #14
ok. i think the whole f(x) and f(1/x) thing was just confusing me. i replaced them with x and y and did it like robphy advised and got an answer that the back of the book said was correct.

2x-3y=4
2y-3x=1/4

y=(3x)/2-1/8

2x-3((3x)/2-1/8)=4
2x-(9x)/2 - 3/8=4
(4x)/2-(9x)/2=4+3/8
5x=8+6/8
5x=-64/8-6/8
5x=-70/8
x=-70/40
x=-7/4

so i think f(2)=-7/4.
did i do it right?
 
  • #15
Yeah. Thats it.
 
  • #16
thanks for all the help guys. this question for some reason became an itch for me. i had to solve it. hopefully as i learn more i will be able to better grasp the concepts.
 
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