Consider a linear function  $f\left(x\right)=ax+b$ such that     $f\left(x\right)-g\left(x\right)=44$ for all real values of $x$,where  $g\left(x\right)=f^{-1}\left(x\right)$ for all real  values of $x$.(Note : $f^{-1}$ is the inverse function of f). Then the number of such possible functions f(x) is _______

Since f (x) = ax + b, we can determine an expression for $g\left(x\right)=f^{-1}\left(x\right)\text{ by letting }y=f\left(x\right)$ and to obtain y = ax + b. We then interchange x and y to obtain x = ay + b which we solve for y to obtain ay = x – b or $y=\frac{x}{a}-\frac{b}{a}$ Therefore, $f^{-1}\left(x\right)=\frac{x}{a}-\frac{b}{a}$ 
Note that  $a\ne 0$ (This makes sence since the function f (x) = b has a graph which is a
horizontal line, and so cannot be invertible.)
Therefore, the equation $f\left(x\right)-g\left(x\right)=44$ becomes $(ax+b)-\left(\frac{x}{a}-\frac{b}{a}\right)=44$  or $\left(a-\frac{1}{a}\right)x+\left(b+\frac{b}{a}\right)=0x+44$ is true for all x, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.
Therefore, $a-\frac{1}{a}=0\text{ and }b+\frac{b}{a}=44\text{. }$
From the first of these equations, we obtain $a=\frac{1}{a}\text{ or }{a}^2=1$ which gives
a = 1 or a = -1.
If a = 1, the equation $b+\frac{b}{d}=44$ becomes
$b+b=44$, which gives b = 22.
If a = -1 , the equation $b+\frac{b}{a}=44$ becomes
b – b = 44, which is not possible.
Therefore, we must have a = 1 and b = 22, and so f(x)= x + 22.