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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.$f(x) = \sqrt{4x +3}$

## $y=f(x)=\sqrt{4 x+3}(y \geq 0) \Rightarrow y^{2}=4 x+3 \Rightarrow x=\frac{y^{2}-3}{4}$. Interchange $x$ and $y$: $y=\frac{x^{2}-3}{4} \cdot \operatorname{So} f^{-1}(x)=\frac{x^{2}-3}{4}(x \geq 0) .$ From the graph, we see that $f$ and $f^{-1}$ are reflections about the line $y=x$.

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##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

all right to start this problem, we have function f of X, and we're going to find its inverse. And I think it will be convenient if we call it why, instead of f of X, So to find the inverse, the first thing we want to do is switch X and y. So now we have X equals the square root of four y plus three and then we want to solve for y. So we're going to square both sides, and then we're going to subtract three from both sides, and then we'll divide by four. Now, let's call this inverse f inverse instead of calling it. Why so f inverse of X equals X squared minus 3/4. Okay, so now we grab a calculator and we want to graph what we just found. F enders X squared minus 3/4, as well as the original function F for X plus three as well as the line y equals X. And I'm going to use a decimal window zoom decimal, which has a nice spread of equal spacing for X and y axes. Okay, so here's what we're looking at. So if you pressed Trace you can tell which graph is which. Now the blue one is the inverse. That was the parabola. Now we need to ignore part of it. We don't actually want the full parabola because the square root function was not a full proble. It was only half so we only want the right side. Basically. So what we could Dio is go back to our graph and imagine that we're just getting rid of this extra piece here. We don't want any of that. It goes to Oh, roughly here. Okay, that's him out. We want now as we look at it. What's the relationship that we notice? Do we see that these graphs are reflections about the line Y equals X? They definitely look like it.

Oregon State University
##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp