Help with finding inverse. Calculus. Problem I\'m working with below, having tro

Question

Help with finding inverse. Calculus.

Problem I'm working with below, having trouble understanding it. From the problem below I don't understand where 121b^2 comes from and could use help explaining that as the problem doesn't explaing it well.

A formula for a function y = f(x) is given below. Find f-1 (x) and identify the domain and range off-1 As a check, show that f (f-1 (x)) = f-1 (f(x)) f(x) x2 - 22bx, b 0 and constant, xs b The process of passing from f to f-1 can be summarized as a two-step procedure. First solve the equation y = f(x) for x This gives the formula x f-1(y) where x is expressed as a function of y. Then interchange x and y, obtaining a formula y = f-1(x) where f-1 is expressed in the conventional format with x as the independent variable and y as the dependent variable. Solve for x in terms of y by completing the square. f(x) x2-225x 2 y = x-22DX y + 121b2-x2-22DX + 121b2

Explanation / Answer

Here 121b^2 is taking for just simplification of the equation y=x^2 - 22bx. That is ; it is not coming from anywhere. We just add it both side because if we add this in both side then the right hand side can be made a square of (x - 11b).That is right hand side is made (x - 11b)^2.Then we solve the equation, y =x^2 - 22bx very easily.

Now you see how to slove x in terms of y.

Now

y +121b^2 =(x - 11b)^2 implies x= 11b + squart root of (y+ 121b^2). That we express x in terms of y. Then you can set inverse of f easily. Actually the inverse is 11b + square root of (x+ 121b^2).

I hope you understand it.

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