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Find an explicit formula for $ f^{-1} $ and use it to graph $ f^{-1} $, $ f $, and the line $ y = x $ on the same screen. To check your work, see whether the graphs of $ f $ and $ f^{-1} $ are reflections about the line.

$ f(x) = 1 + e^{-x} $

see solution

01:23

Jeffrey P.

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

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all right. In this problem, we have f of X, and what we want to do is find its inverse. So let's go ahead and change f of x toe. Why? So we have y equals one plus each of the negative X and then to find the inverse, we want to switch positions of X and y. So now we have X equals one plus each of the negative. Why? And now we're going to sell for why So let's start by subtracting one from both sides, and then we'll take the natural log of both sides that will leave negative wide by itself. And then we'll multiply both sides by negative one to get why by itself. And then instead of calling it why we can call it f inverse of X. So what we're going to do is graph this graph the original f of X and graph the line y equals X and look and see if the two functions are reflections across the line Y equals X. So we grab a calculator, press y equals. We have our original function f of X here, and we have our inverse that we just found and we have y equals X, and I'm going to look at the's in a zoom decimal window so that zoom number four. So the question is, do they look like reflections across the line? Y equals X? Definitely they dio.

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