Math Help - Stuck on transformation question

1. Stuck on transformation question

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

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

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

My effort:

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

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

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

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

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

What have I done wrong?

2. Re: Stuck on transformation question

Originally Posted by angypangy
$f(x) = \frac{e^{2x}}{x^2}$

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

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

My effort:

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

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

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

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

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

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

What have I done wrong?
I get the same answer as you, I have put my working in a spoiler.

Spoiler:
$f(x-1) = \dfrac{e^{2x}}{e^2(x-1)^2} = \dfrac{e^{2x}}{e^2} \cdot \left(\dfrac{1}{x-1}\right)^2$

From the question: $f(x-1) = k \left(\dfrac{1}{x-1}\right)^2 f(x)$

$\dfrac{e^{2x}}{e^2} \cdot \left(\dfrac{1}{x-1}\right)^2 = k \left(\dfrac{1}{x-1}\right)^2 \cdot \dfrac{e^{2x}}{x^2}$

Immediately we can see that $e^{2x} \text{ and } \left(\dfrac{1}{x-1}\right)^2$ cancel:

$\dfrac{1}{e^2} = \dfrac{k}{x^2}$

$k = \dfrac{x^2}{e^2}$

3. Re: Stuck on transformation question

Originally Posted by angypangy
$f(x) = \frac{e^{2x}}{x^2}$

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

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

My effort:

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

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

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

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

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

What have I done wrong?

$if~f(x)=\frac{{e}^{2x}}{(x)^2}~~~then~\\ \\ f(x-1)=\frac{{e}^{2(x-1)}}{(x-1)^2}~=~ \frac{{e}^{2x-2}}{(x-1)^2}~\\ ~~~~~~~~~~~~=\frac{e^{2x}.e^{-2}}{(x-1)^2}~=~ e^{-2}.\frac{{e}^{2x}}{(x-1)^2}~~\\ ~~~~~~~~~~~~=\frac{1}{{e}^2}\frac{{e}^{2x}}{(x-1)^2}~=~\frac{1}{{e}^2}.f(x)$

4. Re: Stuck on transformation question

vernal,

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

5. Re: Stuck on transformation question

Originally Posted by Deveno
vernal,

$\frac{e^{2x}}{(x-1)^2} \neq f(x) = \frac{e^{2x}}{x^2}$
oh sory

$ohh~~ sory\\ ~~~~~~~~~~~~\frac{1}{{e}^2}\frac{{e}^{2x}}{(x-1)^2}~=~\frac{1}{{e}^2}.\frac{x^2}{(x-1)^2}.\frac{{e}^{2x}}{x^2}~=~\frac{1}{{e}^2}.\frac{x^2}{(x-1)^2}.f(x)~=~ \frac{x^2}{e^2}.{(\frac{1}{x-1})}^2.f(x)$

6. Re: Stuck on transformation question

Wow, I beat the book That feels good!

7. Re: Stuck on transformation question

yes, and that leads to the same result as the OP and e^(i*pi) obtained, that $k = \frac{x^2}{e^2}$