Originally Posted by

**angypangy** $\displaystyle f(x) = \frac{e^{2x}}{x^2}$

Question is, show that, for a certain value of k,

$\displaystyle f(x - 1) = k(\frac{1}{x-1})^2 f(x)$

My effort:

$\displaystyle f(x-1) = \frac{e^{2(x-1)}}{(x-1)^2}$

$\displaystyle = \frac{e^{2x-2}}{(x-1)^2}$

$\displaystyle = \frac{e^{2x}}{e^2(x-1)^2}$

$\displaystyle = \frac{x^2}{e^2} (\frac{1}{x-1})^2 \frac{e^{2x}}{x^2}$ .

Therefore k = $\displaystyle \frac{x^2}{e^2}$

But apparently the answer should be $\displaystyle \frac{1}{e^2}$

What have I done wrong?