In Exercises 39-52,
a. Find an equation for ƒ¯¹(x).
b. Graph ƒ an...

Question

In Exercises 39-52,
a. Find an equation for ƒ¯¹(x).
b. Graph ƒ and ƒ¯¹(x) in the same rectangular coordinate system.
c. Use interval notation to give the domain and the range off and ƒ¯¹. f(x) = √(x-1)

Accepted Answer

Welcome back. I am so glad you're here. We are given this function, we're given F of X, which is equal to the square root of an inside the radical X minus 11. We have to do several things, we have to find an equation for its inverse which will be f inverse of X. We need to graph both the function and its inverse. I'm going to draw a quick coordinate plane with a vertical Y axis, horizontal X axis. They come together at the origin in the middle And then we need to provide the domain and range of both functions in interval notation. So lots to do. Let's start with finding the equation for our functions inverse. Going to start off by writing our function f of x equals the square root of X -11. And now we're going to replace F of X with the Y. They're the same thing, so Y equals the square root of x minus 11. Now we're going to kind of enter into the twilight zone here anywhere we see a Y. We're going to replace it with an X. Instead of Y equals its X equals. And anywhere we see an X, we're going to replace it with a Y. So instead of the square root of X minus 11, it's the square root of why And still minus 11. And now we need to get that. Why by itself. Why is not happy until is all by itself on one side? So to get that by itself will square both sides to get rid of that square root. So we'll have X squared on the left equal to y minus on the right to get y by itself will add 11 to both sides. So on the left will have X squared plus 11 and that equals why? Why is happy now? This is wonderful. And this X squared plus 11. That is the inverse of our function. That's the formula. So f inverse of X equals X squared plus 11. Okay, we found that equation. Now let's get to work on graphing. Let's start with our affects. So if you know exactly what The square root of X -11 looks like. That's wonderful. If you don't you can always choose some X values and plug them into the original function. Figure out the corresponding y values and plot those. But before you plot or figure out those X values, you need to see if there are any domain restrictions in the original function. And yeah, everything underneath the radical needs to be greater than or equal to zero. So we need to set X -11 greater than or equal to zero. So we'll add 11 to both sides to see what our X values have to be X has to be greater than or equal to 11. So not what I would have thought if I hadn't figured out the domain. So I'm going to choose 11 for an X value 12 and 13 for X values. Now let's plug those into the original function. So we've got the square root of no longer X. But now 11 minus 11? Well 11 minus 11 is zero. So we have the squirt of zero which is zero. Okay, there's one point now let's plug in 12. We have a squirt of 12 minus 11. 12 minus 11 equals one. We have the square root of one which is one. Okay, I see what's happening here instead of 13. I'm going to choose 15. So we have the squared of 15 minus 11. 15 minus 11 equals four. So the square root of four equals 2. 15 to a little bit easier to plot than 13. And then the square to something. So now we can pull out these points looking at the coordinate plane starting at the origin for our 1st 130.11 0. We're going to go to the right 11 spaces on the X axis and up zero spaces for our Y value. So we have an X intercept at 11 0. Now for the 00.12 one starting at the origin going to the right 12 spaces for our X value and up one space for our Y value approximately there now for 15 to starting at the origin going to the right 15 units for our X value and up two units for our corresponding Y value. So starting at starting at 11 0 we're going to go up like this toward positive infinity for both our X and Y axes and there is our F of X. All right. I'll label that. We've got one of them graft one more to go. Now remember we had to discern the domain restrictions for F of X. Before we could choose our X values to throw in to our function. We have to figure out the domain for our inverse to before we can choose our X values. And the domain for our inverse is going to come from the domain and range of our original function. So, let's see how that works. Let's figure out the domain and range of our original function. Okay, The domain is all the possible X values. Well, that starts at 11. Heads toward positive infinity. So we put a bracket to say including 11 and we say 11 comma all the way up to positive infinity. And a parentheses for infinity. Now, for our range, that's going to be all the possible Y values. That starts at zero inclusive of zero. All the way up to positive infinity. So, bracket zero comma infinity parentheses for the range. Alright. Now, for the inverse domain and range here's how this works. The domain of our inverse function is the range of our original function. So the domain for our inverse is going to be zero comma infinity. And very similarly the range for our inverse function is going to be the domain of the original function. So that will be inclusive of 11. All the way up to infinity. Okay. So for our domain we just have to pick numbers zero or greater. They can't be negative. So I'm going to choose 01 and two or X values. Let's plug those in. So for X equals zero. That's zero squared plus 11. 0 square to zero. So this is 11 for one for our X value, that's one squared plus 11. 1 squared is one. So this one plus 11 is 12 for our X value of 22 squared plus 11, 2 squared is four. So we've got four plus 11 that equals 15. Will you look at that for our first point here? 0 11 that's very similar to our 110.11. 0 and 1 12 is very similar to 12. 1 to 15 is very similar to 15 to so you can see that the inverse is just switching the X and Y values. So let's see what that looks like when we plotted on the graph. So starting at the Origin we're going to plot the 150.0 11. We start at the origin go nowhere for our X values and we go up spaces for our Y value up 11 units. So we've got a Y intercept at 0 11. Alright. Now for 1 12 we go from the origin one to the right for our X value and up 12 for our Y value approximately here. And for 2 15 we started the Origin go to the right to for our X value and up 15 for our Y. Value. That's getting up into nowhere territory. So drawing from 0 11, it's heading toward positive infinity for X. And Y values. This is our f inverse of X. And you can note that this reflects over the line. Y equals X. So cool. All right now we have found the equation graft both functions found the domain and range for both. Now we just gotta find the matching answer choice. Let's highlight where this F. Of X hits the X. Axis at 11 0. Let's look for a corresponding graph that does that. So for answer choice A. The F. Of X hits the X. Axis at negative 11 0, nope. That's not gonna work. Answer choice A. You're rejected. Alright, let's look at answer choice B. That one hits the X axis at 11 0. Cool. How about the inverse? Well does that hit the Y axis at 0. 11? Yes. It does be. You've got potential. It could be you let's go see what's happening with C. And D. Real quick. Come with us graph. You can come too. Okay, so F. Of X. Hitting the X. Axis at 11 0. No, it hits the Y. Axis at 0. 11. For answer choice. C. You're rejected. And for answer choice. D. Do you hit the Y. Or the X. Axis at 11 0. No. You hit the Y axis at zero negative 11. You are also rejected answer choice D. So it looks like B is going to be the winner. Let's just go check the rest of our information against answer choice B. If we can fit everything on one screen. Okay. Answer choice B. You've got a formula F inverse of X equals X squared plus 11. That's right. That's what we got. Alright. Domain of F of X is 11 to infinity. Yes. Range of F of X is zero to infinity. Yes. Domain of F inverse of X. Zero to infinity. Yes. Range of F inverse of X is 11 to infinity. Yes. And our graph matches answer choice B win. Alright, well done. You've done it, we'll catch you on the next one.

In Exercises 39-52, a. Find an equation for ƒ¯¹(x). b. Graph ƒ and ƒ¯¹(x) in the same rectangular coordinate system. c. Use interval notation to give the domain and the range off and ƒ¯¹. f(x) = √(x-1)

Key Concepts

Inverse Functions

Domain and Range

Graphing Functions

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