, 2020

Solution 1

Since $f(x) = ax+b$ for all real numbers $x$, then $f(t) = at+b$ for some real number $t$.
When $t = bx+a$, we obtain $f(bx+a) = a(bx+a) + b = abx + (a^2+b)$.
We also know that $f(bx+a) = x$ for all real numbers $x$.
This means that $abx + (a^2 + b) = x$ for all real numbers $x$ and so $(ab-1)x + (a^2+b) = 0$ for all real numbers $x$.
For this to be true, it must be the case that $ab=1$ and $a^2 + b = 0$.
From the second equation $b = -a^2$ which gives $a(-a^2) = 1$ and so $a^3 = - 1$, which means that $a=-1$.
Since $b = -a^2$, then $b = -1$ as well, which gives $a+b=-2$.

Solution 2

Since $f(x) = ax+b$ for all $x$, then when $x=a$, we obtain $f(a) = a^2 + b$.
Since $f(bx+a) = x$ for all $x$, then when $x=0$, we obtain $f(a) = 0$.
Comparing values for $f(a)$, we obtain $a^2+b=0$ or $b = -a^2$.
This gives $f(x) = ax - a^2$ for all real numbers $x$ and $f(-a^2x+a) = x$ for all real numbers $x$.
Since $f(-a^2x+a) = x$ for all $x$, then when $x=-1$, we obtain $f(a^2+a)=-1$.
Since $f(x) = ax - a^2$ for all $x$, then when $x=a^2+a$, we obtain $f(a^2+a) = a(a^2+a) - a^2$.
Comparing values for $f(a^2+a)$, we obtain $a(a^2+a) - a^2 = -1$ or $a^3 = -1$.
Since $a$ is a real number, then $a=-1$.
Since $b=-a^2$, then $b=-1$, which gives $a+b=-2$.
Checking, we see that if $f(x)=-x-1$, then $f(-x-1) = -(-x-1)-1 = x$, as required.