Given the function $f\left(\frac{2x+5}{2x-5}\right)$, how do we find the function $f\left(-\frac{5x-2}{5x+2}\right)$?

Question

If $$f\left(\frac{2x+5}{2x-5}\right)=\frac{x+1}{2x+3}\qquad\text{and}\qquad f\left(-\frac{5x-2}{5x+2}\right)=\frac{ax+b}{cx+d}$$ The value of $a+b+c+d$ is equal to:

$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)4$

To solve this problem I realized that the first input of function is related to the second input: Assume that $$\frac{2x+5}{2x-5}=\frac{ex+f}{gx+h}$$ Then the second input of function is: $$-\frac{5x-2}{5x+2}=\frac{hx+g}{fx+e}$$ So we see the two numbers in the same diagonal switched. But I don't know how this fact help me to solve the problem.

Answer 1

My other answer made me realize this more quick approach. Note that substituting $x\mapsto-\frac{1}{x}$ gives:

$$\begin{align} f\left(\frac{2(-1/x)+5}{2(-1/x)-5}\right) &=\frac{(-1/x)+1}{2(-1/x)+3}\\ f\left(\frac{-2+5x}{-2-5x}\right) &=\frac{-1+x}{-2+3x}\\ f\left(-\frac{5x-2}{5x+2}\right) &=\frac{x-1}{3x-2}\\ \end{align}$$

Answer 2

HINT

I would start with noticing that

\begin{align*} f\left(\frac{2x+5}{2x-5}\right) & = f\left(1 + \frac{10}{2x-5}\right)\\\\ & = \frac{x+1}{2x+3}\\\\ & = \frac{1}{2}\left(\frac{2x + 2}{2x + 3}\right)\\\\ & = \frac{1}{2}\left[\frac{(2x - 5) + 7}{(2x- 5) + 8}\right] \end{align*}

Hence the following relation holds: \begin{align*} f\left(1 + \frac{10}{y}\right) = \frac{1}{2}\left(\frac{y+7}{y + 8}\right) \end{align*}

Finally, we can set the substitution: \begin{align*} y = \frac{10}{u - 1} \Rightarrow f(u) = \frac{3+7u}{4+16u} \end{align*}

Can you take it from here?

Answer 3

The question is flawed because $\frac{ax+b}{cx+d}=\frac{kax+kb}{kcx+kd}$. So $a+b+c+d$ does not equal any one thing. But I will assume they should be integers, reduced to have no nontrivial common divisor.

If $g(x)=\frac{2x+5}{2x-5}$, then $g$ is a fractional linear transformation. You can represent it as $g:\begin{bmatrix}x\\1\end{bmatrix}\mapsto\begin{bmatrix}2&5\\2&-5\end{bmatrix}\begin{bmatrix}x\\1\end{bmatrix}$. Here we identify the real number $\frac{a}{b}$ with all scalar multiples of the vector $\begin{bmatrix}a\\b\end{bmatrix}$.

Composition of such functions corresponds to multiplying the corresponding matrices.

So it seems $f$ is also a fractional linear transformation, and $$M_f\cdot \begin{bmatrix}2&5\\2&-5\end{bmatrix}=\begin{bmatrix}1&1\\2&3\end{bmatrix}$$

And you are asked to find what is $$M_f\cdot \begin{bmatrix}-5&2\\5&2\end{bmatrix}$$

Your observation about how the coefficients relates to each other is this, in matrices:

$$\begin{bmatrix}2&5\\2&-5\end{bmatrix}\begin{bmatrix}0&1\\-1&0\end{bmatrix}=\begin{bmatrix}-5&2\\5&2\end{bmatrix}$$

So:

$$\begin{align} M_f\cdot \begin{bmatrix}2&5\\2&-5\end{bmatrix} &=\begin{bmatrix}1&1 \\ 2&3\end{bmatrix}\\ M_f\cdot \begin{bmatrix}2&5\\2&-5\end{bmatrix}\begin{bmatrix}0&1\\-1&0\end{bmatrix} &=\begin{bmatrix}1&1 \\ 2&3\end{bmatrix}\begin{bmatrix}0&1\\-1&0\end{bmatrix}\\ M_f\cdot \begin{bmatrix}-5&2\\5&2\end{bmatrix} &=\begin{bmatrix}-1&1\\-3&2\end{bmatrix}\\ \end{align}$$

So you have $$\frac{ax+b}{cx+d}=\frac{-x+1}{-3x+2}$$

And $a+b+c+d=1$.

Answer 4

Standard way to solve such questions (brute force, no tricks):

Let $\dfrac{2x+5}{2x-5} =t \implies x= \dfrac{5t+5}{2t-2} \implies f(t)= \dfrac{x+1}{2x+3}=\dfrac{\dfrac{5t+5}{2t-2}+1}{2*\dfrac{5t+5}{2t-2}+3} = \dfrac{7t+3}{16t+4}$

$$\implies f(x) = \dfrac{7x+3}{16x+4}$$

$f\left(\dfrac{2-5x}{2+5x}\right)= \dfrac{7\left(\dfrac{2-5x}{2+5x}\right)+3}{16\left(\dfrac{2-5x}{2+5x}\right)+4} = \dfrac{-20x+20}{-60x+40} =\dfrac{-x+1}{-3x+2} $

So question is wrong... it must say: $a,b,c,d$ must be minimized with coeeficent of x positive because the answer can also be $-20+20 -60+40 = -20$ etc etc.