Let \(f(x)=\frac{x}{x-1}\). Then there exist unique real numbers \(p, q,\) and \(r\) such that

\(f(5x)=\frac{pf(x)+q}{rf(x)+1}.\)

Find \(p + q + r\).

proyaop Oct 4, 2022

#1**+3 **

Hi proyaop!

\(f(x)=\dfrac{x}{x-1}\)

So: \(f(5x)=\dfrac{5x}{5x-1}\)

Now equate the given with this:

That is,

\(\dfrac{5x}{5x-1}=\dfrac{pf(x)+q}{rf(x)+1}\)

But, we know \(f(x)\):

\(\dfrac{5x}{5x-1}=\dfrac{p(\dfrac{x}{x-1})+q}{r(\dfrac{x}{x-1})+1}\)

Multiply the right with (x-1)/(x-1):

\(\dfrac{5x}{5x-1}=\dfrac{p*x+q(x-1)}{r*x+(x-1)}\)

Expand and equate.

\(\dfrac{5x}{5x-1}=\dfrac{(p+q)x-q}{(r+1)x-1}\)

Well by comparing we get:

\(p+q=5 \\ q=0 \\ r+1=5 \\\)

Hence, \(p=5,q=0,r=4\)

So, \(p+q+r=5+0+4=9\)

I hope this helps!

Guest Oct 5, 2022

#2**+2 **

I got it! **The answer is 9!**

How I did it:

First I plugged in x as 0, f(0) = 0 / -1 = 0

(0p + q)/(0r + 1) = f(5 * 0) = 0 => q/1 = 0 => **q = 0**

Next I plugged in x as -1, f(-1) = 1/2

(p/2 + 0)/(r/2 + 1) = f(-5) = 5/6

p/(r + 2) = 5/6

__6p = 5r + 10 (1)__

Next I plugged in x as -2, f(-2) = 2/3

(2p/3)/(2r/3 + 1) = f(-10) = 10/11

2p/(2r + 3) = 10/11

22p = 20r + 30

__11p = 10r + 15 (2)__

Take equation (1) and multiply by 2 => __12p = 10r + 20 (1b)__

Take equation 1b - equation 2 => **p = 5**

Plug it back into equation 1 => 30 = 5r + 10 => **r = 4**

p + q + r = 5 + 0 + 4 = 9

proyaop Oct 5, 2022