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Find a formula for the inverse of the function.

$ y = \dfrac{1 - e^{-x}}{1 + e^{-x}} $

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$f^{-1}(x)=\ln \left(\frac{1+x}{1-x}\right)$

04:41

Jeffrey Payo

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 5

Inverse Functions and Logarithms

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Johns Hopkins University

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

Lectures

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A multivariate function is a function whose value depends on several variables. In contrast, a univariate function is a function whose value depends on only one variable. A multivariate function is also called a multivariate expression, a multivariate polynomial, a multivariate series, or a multivariate function of several variables.

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In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Find a formula for the inv…

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Find the inverse of$$\…

Let's find the inverse. So we start by switching X and Y, and we get X equals one minus e to the negative. Why over one plus e to the negative y. Now our goal is to isolate. Why. So how about if we multiply both sides by one plus e to the negative Y? That will give us X times a quantity one plus each of the negative. Why equals one minus each of the negative way? Okay, now let's distribute the X and we have X Times or X plus X E to the negative Y equals one minus e to the negative way. Now let's see if we can get those white terms on the same side. So I'm going to add E to the negative white of both sides. And then I'm going to get the other terms on the other side. So I'm going to subtract X from both sides. Now it's factor E to the negative. Why, out of both terms on the left, So we've e to the negative y times one plus x equals one minus X. Now we can divide both sides by one plus x. So if e to the negative y equals one minus X over one plus X. Now it's time for logarithms to get the Y out of the exponents. We need logarithms. So let's take the natural log of both sides and on the left were just going to be left with negative. Why? Because the natural log of E to the negative why is just going to be negative. Why? And that will equal the natural log of one minus X over one plus X. Now let's go ahead and multiply both sides by negative one, and we get we get Why equals that negative that sorry y equals the opposite of the natural log of one minus X over one plus X. Now this answer is perfectly fine. There's nothing wrong with it, but there is a way to simplify it a little bit further. So if you have this answer and you looked it up and the answer in the back of the book look different, you would want to know why. So we can use the power property of logarithms and bring the negative one up and make it a power so that gives us why equals the natural log of one minus X over one plus X rays to the negative one power and what is a negative one? Power. It's a reciprocal. And so if we take the reciprocal of the quotient we have, why equals natural log of one plus X over one minus X? Now we could call this F inverse if we want to, so we have finally f inverse of X equals the natural log of one plus X over one minus x.

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