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If $f(x)=\frac{2 x+1}{3 x+2},$ find $\left(\text { a) } f^{-1}(0), \text { (b) } f^{-1}(-1)\right.$

(a) $-1 / 2$(b) $-3 / 5$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 1

Inverse Functions

Harvey Mudd College

Baylor University

Idaho State University

Lectures

00:04

If $f(x)=x^{2}+1,$ find

03:05

If $f(x)=\frac{2 x-3}{3 x+…

01:46

If $f(x)=x^{3}+x^{2}-x-1,$…

01:14

If $f(x)=\left\{\begin{arr…

01:52

$$\text { If } f(x)=1+…

01:48

Let $f(x)=\frac{2 x+1}{x^{…

01:03

If $f(x)=\left\lfloor x-1 …

01:26

If $f(x)=|x-1|+\left\lflo…

02:52

If $ f( x) = x^2/ (1 + x),…

01:07

If $f(x)=A x^{2}-3 x$ and …

so before we actually try to find these in versus, um, we should look at the range of dysfunction here just to make sure that these inputs are going to be defined. And if you were to find what the range of this is, um, using the fact that this is a rational expression, uh, it should end up just being from negative infinity to two thirds union two thirds to infinity, which remember, then tells us the domain of F inverse is going to be that same thing. So this is negative. Infinity 22 3rd union, two thirds to infinity. So both of these values we can plug in so they should both be defined. Uh, and the reason why we want to do this is because sometimes when you're just like moving things around, uh, sometimes things just kind of get lost in translation. So it's good to just kind of keep that in mind. Um, yeah, So now let's just go ahead and solve for the inverse of this, and then we can just go and plug those in. So first we change the values around. So it would be X is equal to two y plus 1/3 Y plus two, and then we're going to solve for X. So the first thing we're gonna do is clear our denominator. So maybe three x y plus two x is equal to two y plus one. So I'm gonna get my, uh my wife is on the left. My ex is on the right to be three xy minus two. Y is equal to ah, one minus two X. We can go ahead and factor out the Y here and then divide this over so we get Y is equal to one minus two X over three X minus two and so this is going to be our inverse. So let me actually scoot this here. And so now we just need to plug in zero, plug a negative one, and that would be on our way. So I'll just do these in blue. So f inverse is going to be one minus 0/0 minus two. So that would be negative one half and then down here. If we plug in negative one, that would be one plus two over negative three minus two, which would be three over negative five. So those are going to be what we get when we evaluate it at the inverse

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