+1632 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 \(p + q + r\).
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!
+1632 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
