+0

# Hard function problem --- help!

+2
180
4
+1315

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 $$p + q + r$$.

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!

Oct 5, 2022
#3
+1315
+1

Thanks guest, you have a much smarter way of algebraically solving it --- I just plugged in random values of x and hoped for the best...

proyaop  Oct 5, 2022
#4
+1

Actually, your idea also is pretty neat! I learnt something from it :D!

Guest Oct 5, 2022
#2
+1315
+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

Oct 5, 2022