1. The statement of this problem is not very clear. Is k a constant? or k is another function and g(x) =(kf)(x), the product of the functions k and f? If k is a constant, then the only value of k that makes g(x)=kf(x) not invertible is k = 0, and it makes no sense to ask for the sum of all possible values of k since there is only one value, that is 0. If k is a nonzero constant, then g always has an inverse; in fact \(g^{-1}(x)=f^{-1}(\frac{x}{k})\). For instance, we know that if \(f(x)=x^3\), then \(f^{-1}(x)=\sqrt[3] {x}\). The inverse of \(g(x)=kx^3\) is just \(g^{-1}(x)= \sqrt [3]{\frac{x}{k} }\), assuming k is not equal to zero.
If on the other hand k is a function and g(x) is the product of f and k, again the question makes no sense since there are lots of functions that when multiplied by f give a function that is not invertible. Two examples: \(g(x)=x^3\cdot \frac{1}{x} \), where \(f(x)=x^3\ and \ k(x)= \frac{1}{x} \); and \(g(x)=x^3 \cdot \frac{x+1}{x} \), where \(f(x)=x^3\) and \(k(x)= \frac{x+1}{x} \). In both cases g is not invertible because it is not 1-1.
2. Here I will assume you mean \(f(x)= \frac{ax+b}{cx+d} \). The answer to this question is: \(a+d=0\). There is quite a bit of Algebra involved in showing this and it starts by setting \(f(f(x))=x\), which means f is its own inverse. We have
\(f(f(x))=f(\frac{ax+b}{cx+d} )= \frac{a(\frac{ax+b}{cx+d})+b }{c(\frac{ax+b}{cx+d})+d } =\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} =\frac{x}{1} \);
If you cross-multiply and combine terms you get,
\((a^2+bc)x+(ab+bd)=(ca+dc)x^2+(cb+d^2)x\).
Setting the coefficients of the polynomials on the two sides of the equation equal, we get
\(ab+bd=0\), or \(b(a+d)=0\),
\(ca+dc=0\), or \(c(a+d)=0\)
\(a^2+bc=bc+d^2\), or \(a^2-d^2=0\).
The first two equations tell us that \(a+d=0\), since neither b nor c could be zero (one of the assumptions is that \(abcd \neq 0\))
