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 F of X, which is equal to the square root of X plus 11. And we need to do several things, we need to find the equation for its inverse which is f inverse of X. We need to graph both our function and its inverse. So I'm going to draw a quick coordinate plane. We've got a horizontal X axis, a vertical Y axis and the meat together at the origin in the middle. And we need to provide the domain and range and interval notation for both our function and its inverse. Okay, so the first thing I want to work on is finding the equation for its inverse to do that. We're going to start with our original function. So we'll start with F of X equals the square root of X plus 11. And we are going to rewrite this F of X as Y equals so Y equals the square root of X plus 11. Now we're going to kind of enter into the twilight zone anywhere. We see why we're going to replace that with an X. So instead of Y equals its X equals. And anywhere we see an X, we're going to replace that with a Y. So instead of the square root of X, it's the square root of Y. Still plus 11 though. And what we want to do is get why by itself, why is only happy when it is just all by itself. So we don't want to get it to one side subtracting 11 from both sides will have x minus 11 equals a squared of y. And then to get rid of that square root we're going to square both sides. So on the right hand side that's why equals. And on the left hand side that's parentheses X minus Squared. And this X -11 squared part, that is the inverse of our original function. So we can write that here. It's parentheses X -11 and parentheses squared. So now we can go back up and work on the next step which is to graph both functions starting with our original function. If you know exactly what the squared of X plus 11 looks like on the graph. Wonderful. If you don't you can choose some X values, plug them into the original function and find the corresponding Y values. Then plot those points before choosing our X values. I need to note the domain restriction here what's inside of the radical. And in this case that's X has to be greater than or equal to zero. So for the X values that I choose, I have to choose positive ones. And because they're going inside that radical, I'm going to choose 41 and zero. So now plugging four in for X instead of the squared of X, it's square to four plus 11. The square to four is two. So we've got two plus 11 to plus 11 is 13. Looking at our coordinate plane. Starting at the origin in the middle we're going to the right four units for our X value and up 13 units for the Y value. So something like that. Now let's plug one in for X. We've got the squirt of one plus 11. The squirt of one is one so one plus 11 which equals on our graph, starting at the origin, moving to the right one unit for our X value and up 12 units for our Y values. Something like that right now plugging zero in for X. So the squared of zero plus 11 equals squared of 00. So zero plus 11 that is 11. So starting at the origin we're not moving to the writer to the left for our X value and we're moving up 11 for our Y value. So that has a Y intercept at 0 11. And it starts here and goes off in this direction it's going toward positive infinity for both the X and the Y axis. And I'll label this F of X. We've got one of them graft are right now. Before we can graph the inverse as I said for F of X, we had to figure out what our domain restrictions were. Before we could figure out the possible X values. We could choose well we need to know our domain restrictions for our inverse to but we actually get the domain restrictions from the domain and range of our original function. So we need to write down the domain And the range of our original function 1st. The domain that's all the possible X values. So it starts at zero and heads toward positive infinity. It includes zero. So we put a bracket to say inclusive of than zero comma all the way up to infinity and parentheses for the infinity. Alright. There's the domain the range that's all the possible Y values. Well it starts at 11 and goes up to positive infinity. So it's inclusive of 11 bracket 11 comma infinity parentheses. Alright, So there's our domain and range for our affects now. How does that relate to our inverse? While the inverse, The domain of our inverse function is going to be the range of our original function. So the domain of our inverse is going to be bracket comma infinity. Now, very similarly, the range for inverse function is going to be the domain of our original function. So bracket zero comma infinity. Alright. There's our domain and range for our inverse. Now we can choose some X values that are within the domain. So it has to be a number 11 or greater. So I'm going to choose 11. 12 and 13. Probably not numbers I would have thought of off the top of my head and plugging those X values in to the function. So we'll have instead of x minus 11, it will be 11 minus 11 squared. 11 minus 11 is 00 squared equals zero. So we've got a point at 11 0. Starting at the origin. We're going to the right 11 units. So about here and then it's not going up or down at all for the Y value. So that's got an X intercept at 11 0. Alright, let's keep going plot a couple more, plugging 12 in for X. So 12 minus 11 squared 12 minus 11 is squared equals one. So from the origin going to the right 12 units and then up one unit for our Y value. That's about here. Alright, now, plugging 13 in 13 minus 11 squared 13 minus 11 is 22 squared is four. All right. Now, plugging starting at the origin, going to the right 13 units for our X value and then going up four units for our Y values approximately here, starting at 11 0. It's going to curve up like this. It's going to head toward positive positive infinity for both our x and Y values. And as you may have noticed all of our X values and our original function correspond with the Y values in our inverse function and they match up with the corresponding inverse of the X and Y values for the original and inverse functions. This reflects over the line X equals Y or y equals X. Is how you would write the equation for that line, Y equals X. So we've got these both plotted, we have figured out the inverse, we have figured out the domain and range. Now we're ready to choose the correct answer choice. So let's highlight let's look for an F of X that hits the Y axis at 0 11. So for answer choice A the F of X hits it hits the X axis at negative 11 0. That does not match. So not you answer choice A. All right. Now let's look at answer choice. B. The F of X hits the X axis at 11 0. That is not what we are looking for. So, sorry, answer choice B. You do not work either. Now let's grab our graph and bring it down to see answer choices C. And D. And we've got at answer choice C. The F of X hits the Y axis at 0 11. That's looking good. The F inverse of X hits the X axis at 11 0. That looks good. Alright, let's really quick look at answer choice D. That one has the F of X hitting at zero negative 11. No, that's no good. So, we can rule out answer choice. D. Let's check the rest of our answer choice C. So we've got a formula for a inverse function and the formula is f inverse of X equals x minus 11 squared, yep. That matches. And we've got domain of F of X equals zero To infinity. That looks good. The range of F of X is 11 infinity. That looks good. All right. The domain of f inverse of X is infinity. That looks good. The range of F inverse of X equals zero infinity. Yes. And the graph matches answer choice C. You have matched. Alright, well done. 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

Graphing Functions

Domain and Range

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