anubis01 said:
No, that's just bad algebra. [(1+e^x)(1+e^x)= 1+ 2ex+ e2x is the numerator and the denominator is 1- e2x, not "2".0
And even if it were correct, ln(A+ B) is not ln(A)+ ln(B).
No, it doesn't. The if y= f(x), then x= f-1 of y so the standard way to find the inverse of y= f(x) is to solve the equation for x. If y= (1+ex)/(1- ex) then (1- ex)y= 1+ ex or y- yex= 1+ ex. y(1+ ex)= y- 1, 1+ ex= (y- 1)/y, ex= (y-1)/y- 1= (y-1)/y- y/y= -1/y. Can you solve that for x?
No, ln(AB) is not A ln(B) ln(ce^(bx))= ln(c)+ ln(bx)= ln(c)+ bx.
Since the "x"s cancel out in what you have, you should be sure it is not right!
The inverse of a function is a new function that reverses the input and output values of the original function. In other words, the input of the original function becomes the output of the inverse function, and vice versa.
To find the inverse of a function, switch the x and y variables and solve for y. This will give you the equation of the inverse function. Keep in mind that not all functions have an inverse, and some may have more than one inverse.
The notation for the inverse of a function is f-1(x), where f(x) is the original function. This notation is read as "f inverse of x".
Yes, there are certain restrictions for finding the inverse of a function. The original function must be a one-to-one function, meaning that each input has a unique output. Additionally, the function must pass the horizontal line test, which means that no horizontal line can intersect the graph of the function more than once.
To graph the inverse of a function, switch the x and y coordinates of each point on the original function's graph. This will give you the graph of the inverse function. Keep in mind that the inverse function will be a reflection of the original function over the line y = x.