[Algebra]
Problem: Determine all linear functions $f(x)=a x+b$ such that if $g(x)=f^{-1}(x)$ for all values of $x$, then $f(x)-g(x)=44$ for all values of $x$. (Note: $f^{-1}$ is the inverse function of $f$.)

Solution: Since $f(x)=a x+b$, we can determine an expression for $g(x)=f^{-1}(x)$ by letting $y=f(x)$ to obtain $y=a x+b$. We then interchange $x$ and $y$ to obtain $x=a y+b$ which we solve for $y$ to obtain $a y=x-b$ or $y=\frac{x}{a}-\frac{b}{a}$.

Therefore, $f^{-1}(x)=\frac{x}{a}-\frac{b}{a}$.

Note that $a \neq 0$. (This makes sense since the function $f(x)=b$ has a graph which is a horizontal line, and so cannot be invertible.)

Therefore, the equation $f(x)-g(x)=44$ becomes $(a x+b)-\left(\frac{x}{a}-\frac{b}{a}\right)=44$ or $\left(a-\frac{1}{a}\right) x+\left(b+\frac{b}{a}\right)=44=0 x+44$, and this equation is true for all $x$.

We can proceed in two ways.

Method \#1: Comparing coefficients

Since the equation

$$
\left(a-\frac{1}{a}\right) x+\left(b+\frac{b}{a}\right)=0 x+44
$$

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, $a-\frac{1}{a}=0$ and $b+\frac{b}{a}=44$.

From the first of these equations, we obtain $a=\frac{1}{a}$ or $a^{2}=1$, which gives $a=1$ or $a=-1$. If $a=1$, the equation $b+\frac{b}{a}=44$ 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 $f(x)=x+22$.

Method \#2: Trying specific values for $x$

Since the equation

$$
\left(a-\frac{1}{a}\right) x+\left(b+\frac{b}{a}\right)=0 x+44
$$

is true for all values of $x$, then it must be true for any specific values of $x$ that we choose.

Choosing $x=0$, we obtain $0+\left(b+\frac{b}{a}\right)=44$ or $b+\frac{b}{a}=44$.

Choosing $x=b$, we obtain $\left(a-\frac{1}{a}\right) b+\left(b+\frac{b}{a}\right)=44$ or $a b+b=44$.

We can rearrange the first of these equations to get $\frac{a b+b}{a}=44$.

Using the second equation, we obtain $\frac{44}{a}=44$ or $a=1$.

Since $a=1$, then $a b+b=44$ gives $2 b=44$ or $b=22$.

Thus, $f(x)=x+22$.

In summary, the only linear function $f$ for which the given equation is true for all $x$ is $f(x)=x+22$.

Final answer: $f(x)=x+22$