fghtffyrdmns said:
I have another question of inverse ln.
Let's say you have -(ln (x)+2)^{1/2}
The domain would just be 0 < x </= e^2
In order that we be able to take the logarithm, we must have x> 0. In order to be able to take the square root, we must have ln(x)+ 2\ge 0 or ln(x)\ge -2. Since ln is an increasing function that is the same as saying x\ge e^{-2}. That is NOT the same as x\le e^2! Since e^{-2}> 0, we have both x> 0 and x\ge e^{-2} just by taking x\ge e^{-2}. That is the domain.
The inverse of it would be x = e^{(-y^2 + 2)}
Be careful about writing x as a function of y. If y= 3x, then x= y/3 is the
same function, not the inverse. The inverse function to y= 3x is y= x/3.
The original function is given by y= -(ln(x)+ 2)^{1/2}. The inverse function is given by x= -(ln(y)+ 2)^{1/2} but we want to write "y= " so we need to solve for y. Squaring both sides gives x^2= ln(y)+ 2 so that ln(y)= x^2- 2 and then y= e^{x^2- 2}. The "-" in the original disappeared when we squared both sides. Stictly speaking, the natural domain of y= e^{x^2- 2} is "all real numbers" but this is NOT that function:
Would the domain of the inverse function just be x < 0? which is the range of the original function.
Yes, exactly! When x= e^{-2} in the original function, y= 0. As x increases without bound, ln(x) increases without bound and so does the square root of ln(x)+ 2. But y is the
negative of that. The range of the original function is "all x< 0" and so the domain is also, just as you say! (A function and its inverse "swap" range and domain just as the "swap" x and y.)
