"Show that a function of the form x maps to (x+a)/(x-1), where x belongs to the real numbers and is not equal to 1, is self-inverse for all values of the constant a except one. State the exceptional value of a."
Can someone explain how I should going about proving/solving this please?
Can someone explain how I should going about proving/solving this please?
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Just go about finding the inverse and the unallowed value for a will be obvious.
To find the inverse, suppose y = (x+a)/(x-1). Now switch y and x and solve for y to get the inverse function:
x = (y+a)/(y-1) ==> x(y-1) = y + a ==> (x-1)y = x+a ==> y = (x+a)/(x-1).
Thus the inverse function is equal to the original function. However, notice that if a = -1, and we were to try the above technique, we would get x = 1, which is not a function.
Geometrically, if a = -1, the function would become a horizonal line, f(x) = 1, which is not a one-to-one function and therefore does not have an inverse.
To find the inverse, suppose y = (x+a)/(x-1). Now switch y and x and solve for y to get the inverse function:
x = (y+a)/(y-1) ==> x(y-1) = y + a ==> (x-1)y = x+a ==> y = (x+a)/(x-1).
Thus the inverse function is equal to the original function. However, notice that if a = -1, and we were to try the above technique, we would get x = 1, which is not a function.
Geometrically, if a = -1, the function would become a horizonal line, f(x) = 1, which is not a one-to-one function and therefore does not have an inverse.