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If $f(x)=\sqrt{3 x+2},$ find (a) $f^{-1}(3),$ (b) $f^{-1}(4)$

(a) $7 / 3$(b) $14 / 3$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 1

Inverse Functions

McMaster University

University of Michigan - Ann Arbor

Lectures

01:48

If $f(x)=\sqrt{2 x-1},$ fi…

02:59

(a) $f(x)=(4+3 x)^{1 / 3}$…

03:05

If $f(x)=\frac{2 x-3}{3 x+…

02:10

Find $f^{\prime}(x)$ if $f…

So before we actually find, um, these values here, one thing we need to keep in mind is making sure that are in Vs is actually going to be defined at three and four. And so the way we can find that out is by looking over here to see if our inverse or with the range of our original function is so, uh, this is just like the square root of some number. We know that the range of F is just going to be 02 infinity, because it's just like shifting and compressing the square function horizontally. And so then that translates into our domain or f inverse being the same thing. So in each of these cases, three and four is in the interval 02 infinity. So we have nothing to worry about. So now we can come over here and just plug and why, and then start solving. And the reason why I wanted to say that is because once we find this in first, we'll see that the inverse is how it's written, will have no restriction, honest domain. So for some reason, we had something that was not an actual output of this, we would end up with an undefined expression over here. Um, yeah. So we don't need to worry about that, though, so let's just go ahead and start by switching these around so we can try to solve for X assault For Why? I mean, so we square each side we have X squared is equal to through I plus two. So we're going to subtract two divide by three, and then this is our inverse. So now we just come up here and plug three and plug forward. So this would be nine minus 2/3, so that would be seven thirds. And then over here, uh, that would be 16 minus 2/3, which would be 14 3rd. So then these would be, um, our solutions.

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