Let f(x) = -2/(x-1) + 2. Show that f(1/x) − f(x) = (2x+2)/(x-1).

Substitute $x$ with $\frac{1}{x}$ in the function $f(x)$:

$$f\left(\frac{1}{x}\right) = \frac{-2}{\frac{1}{x}-1} + 2$$

Combine the terms in the denominator:

$$f\left(\frac{1}{x}\right) = \frac{-2}{\frac{1-x}{x}} + 2$$

Invert and multiply the fraction:

$$f\left(\frac{1}{x}\right) = \frac{-2x}{1-x} + 2$$

Substitute the expressions for $f\left(\frac{1}{x}\right)$ and $f(x)$:

$$f\left(\frac{1}{x}\right) - f(x) = \left(\frac{-2x}{1-x} + 2\right) - \left(\frac{-2}{x-1} + 2\right)$$

Distribute the negative sign and cancel out the constant terms:

$$f\left(\frac{1}{x}\right) - f(x) = \frac{-2x}{1-x} + 2 + \frac{2}{x-1} - 2$$ $$f\left(\frac{1}{x}\right) - f(x) = \frac{-2x}{1-x} + \frac{2}{x-1}$$

Note that $1-x = -(x-1)$. Substitute this into the first term:

$$f\left(\frac{1}{x}\right) - f(x) = \frac{-2x}{-(x-1)} + \frac{2}{x-1}$$ $$f\left(\frac{1}{x}\right) - f(x) = \frac{2x}{x-1} + \frac{2}{x-1}$$

Since the denominators are the same, combine the numerators:

$$f\left(\frac{1}{x}\right) - f(x) = \frac{2x+2}{x-1}$$

Factor out 2 from the numerator:

$$f\left(\frac{1}{x}\right) - f(x) = \frac{2(x+1)}{x-1}$$

This matches the expression we needed to show.