Ytukyg

$$f(ax+b) = x$$
$$f(a) = dfrac{b}{a}$$

I want to determine $f(0)$ from given equations.

My attempt:

$$f(a)= dfrac{b}{a} implies f(0) = dfrac{b}{0} = text{undefined} $$

However, directly determining $f(0)$ makes no sense in this case. Could you assist me with that?

By the way upon computing, the function is $f(x)=dfrac{b(2x-a)}{a^2}$ and $f(0)=dfrac{-b}{a}$. Further using $f(ax+b)=x$ we get $b=dfrac{a}{2}$. So the function is $f(x)=dfrac{2x-a}{2a}$ and $f(0)=dfrac{-1}{2}$, $f(a)=dfrac{1}{2}$.

Well, you want the argument of your given function to be equal to zero. Thus, considering the argument of the funtion $f(ax+b) = x$, we have :

$$ax + b = 0 Leftrightarrow x = -frac{b}{a}$$

Then, $f(0)$ would be yielded for $x= -b/a$, thus again by substituting in the first function expression :

$$f(0)=-frac{b}{a}=-f(a)$$

Here is a start - and contrary to comments, there is a solution.

Let $y=ax+b$ so that $f(y)=x=cfrac {y-b}a$ so now we have expressed $f$ as a linear function of its argument.

I have not yet used the value for $f(a)$ that has been given, so this is the next step.