Equivalence of Two Expressions Involving a Function

[turophile]

The problem:

Let p(x) = 1/x – 1 (x ? 0). Verify that (– 1)/p(x + 1) = p(x) + 2.

I'm trying to work this from the front and from the back. From the front, I have:

(– 1)/[1/(x + 1) – 1] = (– 1)/[(1 – x – 1)/(x + 1)] = (– 1)/[(– x)/(x + 1)] = (x + 1)/x = ...

From the back, I have:

... = (x + 2)/(x + 1) = (1 + x + 1)/(x + 1) = 1/(x + 1) + 1 = [1/(x + 1) – 1] + 2 = p(x) + 2

I'm having trouble connecting the two. How do I get from (x + 1)/x to (x + 2)/(x + 1)? Any hints?

 

[BigGlenntheHeavy]

\(\displaystyle p(x) \ = \ \frac{1}{x}-1 \ = \ \frac{1-x}{x}, \ x \ \ne \ 0.\)

\(\displaystyle p(x+1) \ = \ \frac{1-x-1}{x+1} \ = \ \frac{-x}{x+1}\)

\(\displaystyle \frac{-1}{p(x+1)} \ = \ \frac{x+1}{x} \ = \ p(x)+2 \ = \ \frac{1-x}{x}+2 \ = \ \frac{1-x+2x}{x} \ = \ \frac{x+1}{x}, \ QED\)

 

[turophile]

I see my error now. Sorry I didn't double check my work better before posting. Thanks!