b(ax+b)+a=x, abx+b²+a=x. ab=1, b²+a=0. a=1/b, b²+1/b=0. b isn't 0 or ∞, so b³+1=0. Meaning b is one of the 3 cube roots of -1. Meanwhile, a, being its reciprocal, is b's conjugate. Either a=b=-1, or a=(1±√(3)i)/2 and b=(1-±√(3)i)/2. Meaning either a+b=-2, or a+b=1. The two solutions are -2 and 1.
you can find the inverse of the first one, f(x) = ax + b => f^-1(x) = (x - b)/a. Then equal to the other inverse, (x - b) / a = bx + a x - b = abx + a^2, you find that: x = abx ab = 1, and a^2 = -b. Furthermore, b = -a^2 and replacing in the other one: a(-a^2) = 1 a = -1. Finally, (-1)^2 = - b, b = - 1, so a + b = -1 + (-1) = -2.
I found another way to find the solution. Step 1) Find the inverse f^-1(x) of f(x) while assuming that the given f^-1(x) is different to the actual f^-1(x). Let's call the given f^-1(x) from now on g(x). Well, you'll get f^-1(x) = 1/a * x - b/a as the inverse of f(x) . Step 2) Equate the calculated f^-1(x) and g(x), which means that 1/a * x - b/a has to be equal to bx + a. Step 3) That leads to a system of equations with I) 1/a = b and II) -b/a = a, which is a solvable system of equation. You'll get a = -1 and b = -1. Step 4) Calculate a + b = -1 + (-1) = -2 (sry for my bad english)
b(ax+b)+a=x, abx+b²+a=x. ab=1, b²+a=0. a=1/b, b²+1/b=0. b isn't 0 or ∞, so b³+1=0. Meaning b is one of the 3 cube roots of -1. Meanwhile, a, being its reciprocal, is b's conjugate.
Either a=b=-1, or a=(1±√(3)i)/2 and b=(1-±√(3)i)/2. Meaning either a+b=-2, or a+b=1.
The two solutions are -2 and 1.
you can find the inverse of the first one, f(x) = ax + b => f^-1(x) = (x - b)/a. Then equal to the other inverse, (x - b) / a = bx + a x - b = abx + a^2, you find that: x = abx ab = 1, and a^2 = -b. Furthermore, b = -a^2 and replacing in the other one: a(-a^2) = 1 a = -1. Finally, (-1)^2 = - b, b = - 1, so a + b = -1 + (-1) = -2.
I found another way to find the solution. Step 1) Find the inverse f^-1(x) of f(x) while assuming that the given f^-1(x) is different to the actual f^-1(x). Let's call the given f^-1(x) from now on g(x). Well, you'll get f^-1(x) = 1/a * x - b/a as the inverse of f(x) . Step 2) Equate the calculated f^-1(x) and g(x), which means that 1/a * x - b/a has to be equal to bx + a. Step 3) That leads to a system of equations with I) 1/a = b and II) -b/a = a, which is a solvable system of equation. You'll get a = -1 and b = -1. Step 4) Calculate a + b = -1 + (-1) = -2 (sry for my bad english)
If a and b are not real, you will get three solutions. Sums would be -2, 1, 1.