Question

How do you solve an equation where the function is on both sides?

\[2f(x)-3f(\frac{1}{x}) = x^2\]
Find f(2)

I'm stucked at

\[\frac{2f(\frac{1}{x})}{2} + \frac{x^2}{2}\]

Answer 1

you are trying to find a formula for f correct?

Answer 2

Yes, I meant to write.

f(x) = \frac{2f(\frac{1}{x})}{2} + \frac{x^2}{2}

Answer 3

\[f(x) = \frac{2f(\frac{1}{x})}{2} + \frac{x^2}{2}\]

Answer 4

gimmick is to replace x by 1/x again and get two equations

Answer 5

How would I calculate f(2)?

\[f(2) = \frac{2f(\frac{1}{2})}{2} + \frac{2^2}{2}\]

And

\[f(0.5) = \frac{2f(\frac{1}{0.5})}{2} + \frac{0.5^2}{2}\]

Answer 6

oh wait this is easier than i thought. i thought you were trying to find an expression for
\[f(x)\] but you really just want
\[f(2)\] right? so replace x by 2 and get
\[2f(2)-3f(\frac{1}{2})=4\]

Answer 7

now replace x by 1/2 and get
\[2f(\frac{1}{2})-3f(2)=\frac{1}{4}\]

Answer 8

and you want to solve for
\[f(2)\] so maybe put
\[y=f(2), z = f(\frac{1}{2})\] and solve the system of equations
\[2y-3z=4\]
\[2z-3y=\frac{1}{4}\] for y and z (you will get them both, even though you only need y)
good from there?

Answer 9

Perfect thank you so much!

Answer 10

yw

Answer 11

myininaya is impressed too, i can tell!

Answer 12

to tell you the truth i don't get it

Answer 13

really? you are given the identity
\[2f(x)-3f(\frac{1}{x}) = x^2\] and in fact you can use this to find an expression for
\[f(x)\] but in this case the question was find
\[f(2)\]

Answer 14

so replace x by 2, and the replace x by 1/2, and you get two equations containing f(2)
solve the system and you get it

Answer 15

ok ok
i see it now lol

Answer 16

helps if you have seen a similar problem before. actually i think i answered this here a month or two ago, or at least a the same type of problem

Answer 17

\[2f(2)-3f (\frac{1}{2})=2^2\]

\[2f(\frac{1}{2})-3f(\frac{1}{\frac{1}{2}})=x^2\]

Answer 18

oops i forgot to replace x with 1/2

Answer 19

i have seen system of linear equations
but i haven't seen one like this

Answer 20

gj satellite

Answer 21

try it with x and 1/x instead of 2 and 1/2 and see if you can find the expression for f

Answer 22

me?

Answer 23

why not? i am doing it with scratch paper. bone up on algebra is all it is

Answer 24

\[2f(x)-3f(\frac{1}{x})=x^2\]
\[-3f(x)+2f(\frac{1}{x})=\frac{1}{x^2}\]
\[4f(x)-6f(\frac{1}{x})=2x^2\]
\[-9f(x)+6f(\frac{1}{x})=\frac{3}{x^2}\]
add to get
\[-5f(x)=2x^2+\frac{3}{x^2}=\frac{4x^2+3}{x^2}\] solve to get
\[f(x)=-\frac{4x^2+3}{5x^2}\]

Answer 25

what are these "user study groups"? you know?

Answer 26

\[2 f(x)-3 f(\frac{1}{x})=x^2\]
\[2 f(\frac{1}{x})-3f(\frac{1}{\frac{1}{x}})=(\frac{1}{x})^2 =>2 f(\frac{1}{x})-3f(x)=\frac{1}{x^2} \]

let f(x)=z and let f(1/x)=t

equation 1 is:
\[2z-3t=x^2\]

equation 2 is:
\[2t-3z=\frac{1}{x^2}\]

solve equation 1 for z:
\[z=\frac{x^2+3t}{2}\]

now plug this z into equation 2:
\[2t-3(\frac{x^2+3t}{2})=\frac{1}{x^2}\]

now we solve this for t:
\[2t-\frac{3}{2}(x^2+3t)=\frac{1}{x^2} => (2-\frac{9}{2})t-\frac{3}{2}x^2=\frac{1}{x^2}\]
\[=>\frac{-5}{2}t=\frac{1}{x^2}+\frac{3}{2}x^2\]=>\[t=-\frac{2}{5x^2}-\frac{3}{5}x^2\]

okay so we know t now we will take this t and plug into the expression we have solve for z:
\[z=\frac{x^2+3t}{2}\]
\[=\frac{x^2+3 \cdot (-\frac{2}{5x^2}-\frac{3}{5} x^2)}{2}\]
\[=\frac{x^2}{2}+\frac{3}{2}(-\frac{2}{5x^2})+\frac{3}{2}(\frac{-3}{5}x^2)\]
\[=\frac{x^2}{2}-\frac{3}{5x^2}-\frac{9}{10}x^2\]
\[=-\frac{3}{5x^2}-\frac{4}{10}x^2=-\frac{3}{5x^2}-\frac{2}{5}x^2\]


so recall we let f(x)=z and let f(1/x)=t

this means we have
\[f(x)=-\frac{3}{5x^2}-\frac{2}{5}x^2\]
and
\[f(\frac{1}{x})=-\frac{2}{5x^2}-\frac{3}{5}x^2\]

Answer 27

satellite i don't you combine fractions right

Answer 28

oops i missed the word think

Answer 29

i went the long way

Answer 30

i think i am write. i think

Answer 31

and one reason i think it is that i did it for x = 2, got an answer, and then found
\[f(2)\] and got the same number

Answer 32

oh but i totally wrote the wrong thing! you are right

Answer 33

i wrote
\[f(x)=-\frac{4x^2+3}{5x^2}\] when it should have been
\[f(x)=-\frac{2x^4+3}{5x^2}\]

Answer 34

which is what you have exactly. i claim "typo" as usual