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)², x ≤ 1

Accepted Answer

Welcome back. I am so glad you're here. We are given this function F of X, which equals parentheses x minus five squared. And we're given a domain restriction. We're told that X needs to be less than or equal to five. And we need to do several things with this function. We need to find the equation for its inverse, which is going to be f inverse of X is equal to something. Then we need to graph both the function and its inverse. We can draw a rough sketch of a coordinate plane, will have horizontal X axis of vertical Y axis and they come together at the origin in the middle. Then we need to provide the domain and range in interval notation for both of these. So we have got a lot to do. Leave a little bit of space there. The domain and the range. Okay, let's start off by working on the equation for the inverse of our function. So I'm going to copy down the original functions, equation f of x equals parentheses X - squared. And now I'm going to replace that F of X with a Y. So Y equals parentheses x minus five squared. And now wherever there is a Y, I'm going to replace it with an X. So instead of Y equals its X equals. And wherever there's an X, I'm going to replace it with a Y. So instead of x minus five it's y minus five squared. And now what we need to do is we need to get that. Why by itself. The Y is miserable until it is all alone. So we're going to take the square root of both sides of this equation. And so on the left we have plus or minus the squared of X. And that is equal to why -5. And now wise almost by itself, we're going to add five to both sides of the equation. On the left, we will have five plus or minus the square root of X. And that is equal to why? And this five plus or minus the square root of X. That is the equation for R in verse five plus or minus the square root of X. And we note that in our original function we had a domain restriction where X is less than or equal to five. So we'll need to figure out if this is going to be a five plus the square root of X or five minus the square root of X. That is something that we will need to discern. But right now let's just start graphing our original function. Since we have found the equation for the inverse for the most part. So we'll create an xy table and we'll choose some values for X, plug those into our original function. Figure out the corresponding Y values and then graph those for the X values. Because we have a domain restriction. Our numbers have to be less than or equal to five. I am going to choose 01 and five. Now plugging those into the original equation. So we'll have parentheses instead of X minus five Excuse me that will be zero minus five squared zero minus five is negative five and negative five squared equals positive 25. Now let's plug in one. So we've got parentheses one minus five squared. Well one minus five is negative four, negative four squared is positive 16. Now plugging in 55 minus five squared Well five minus five is 00 squared is zero. So we've got three points. Let's plot them from the origin at 00 for our 00.0 25. We're going to go neither left nor right for our X value of zero. And we're going to go up 25 units for our Y value will mark this as our Y intercept at 0 25 Now for the .116 and we will go to the right one unit from the origin for our x value. And we will go up 16 units for our y value of 16 that's approximately here. Now for the 0.50 we're starting at the origin. We're going to the right five units for our X value and we're going neither up nor down for our Y value. So that is our X intercept at 50 will label that. And then this curves upwards in this direction. Generally speaking and that is our f of X. Alright We've got one of these. Now let's identify the domain and range the domain is all of the possible X values and recall that we have this domain restriction where X has to be less than or equal to five. So the domain can be from negative infinity. We say this in interval notation with parentheses than negative infinity comma. And then it can go up to five including five. And then we put a bracket to say that it does include five. Our range looks like this starts at zero for our Y values and then continues up to positive infinity. So we put a bracket to say including zero comma all the way up to positive infinity and a parentheses for infinity. So there's our domain and range. Now we recall from previous lessons that the domain of our inverse function is going to be the range of our original function. So the domain of the inverse function is going to be inclusive of zero from zero all the way up to positive infinity. And the range for our inverse function similarly that comes from the domain of our original function. So the range of our inverse is going to be from negative infinity. Up to five inclusive of five. So that is very important in helping us discern this plus or minus that we've got here, let's create an xy table And we recall from previous lessons that we should be just swapping our X and Y values to get from our function to its inverse. And so let's take the Y values from our original function will take 25 16 and zero. We'll plug those in for X. And then we'll see what we get for our Y. Values. They should match the X. Values from the original. So we've got two different equations here that we're plugging into. We have five plus the square root of X. And we have five minus the squared of X. Let's plug 25. And for X. So we've got five plus the square root of 25 which is five plus the square of 25 is 55 plus five. That's 10. There's one answer. And now plugging in five minus the square root of X. That's five minus the square root of 25. Well that's five minus the square of 25 is 55 minus five. That equals zero. Well which one fits our range says that we can't get any higher than a positive five. So this 10 doesn't work. We can reject this one. Let's plug 16 in. So we've got five plus the square root of 16. That's five plus the square of 16 is 45 plus four is nine. Now let's plug it into the 2nd 15 minus the square root of 16. That's five minus the square of 16 is 45 minus four, that's one. Well nine is outside of our range. So we reject that one and we accept one and looking back at our original function zero and one are the answers that we're looking for. So we can reject our 10 and are nine and we can accept R. Zero and R. One. Those are within the acceptable range. And we can reject this five plus the square root of X. Our equation is just going to be five minus the square root of X. So now we can plug zero in. We've got five minus the square root of zero. Well the squared of 00. So we've just got five minus 05 minus zero is five. And there's our xy table this matches up and this is our inverse. Now let's plot these three points from the origin 00. We're going to go neither to the left nor to the right for our 05 point. And we are going to go up five units for our Y value. And we can label this this is our Y intercept at 05. Sorry I'm starting at the bottom and now going up for our 0.16 one, we're going to the right 16 units for our X value and we're going up one unit for our corresponding Y value and that can be approximately here and now for the 0.25 zero we're going 25 units to the right of the origin for our X value and we're going neither up nor down for our Y value of zero. So this is not to scale but that's we can call it approximately here and we can label this as being 25 zero for Rx intercept. And now this curves It starts at 05 because of that domain restriction. It can't go any further to the left and this will be our f inverse of X. All right. We can see that we've graphed this and these are inverse ease of each other. They reflect over the line. Y equals X. Yay, we have got our graphing done we have got our equation done. We've got our domains and ranges done. Now let's figure out what the right answer choices. So let's specifically look for we have an F of X with an X intercept at 50. And we have an F inverse of X. With a Y intercept at 05. So, let's look for those specifically. And then we will look for our equation F inverse of X equals five minus the square root of X. Alright, answer choice A. You've got an F of X with an X intercept at negative 50. That is not what we want. So answer choice A. You are rejected. Answer choice B has an F of X. With an X intercept at 50. That's promising. It has an F inverse of X. With a Y intercept at 05. That's promising. It has an equation for the inverse at F inverse of X equals five plus the squared of X. We found out that that does not work. So sorry, answer choice B. You are rejected. Now, let's bring these things down so that we can check out answers C and D. This is going to be a mess for a second while I pull these things down and oops, I lost it. All right, come on down now. We've got to get our domains and ranges. So the domain and range for the original. And our equation of our inverse. Come on down here and our domain and range for the inverse. Okay. Answer choices. C. And D What have you got? Answer choice C has an X intercept for the F of X at 50. Wonderful. It has an A Y intercept for the F inverse of X at 05. Wonderful. It has an equation at F inverse of X equals five minus the square root of X. Yes. That's what we're looking for. The domain of F of X is from negative infinity to five. Yes. The range of F of X is from zero to infinity. Yes. The domain of inverse of X is from zero to infinity. Yes. And the range of F X is from a negative infinity to five. Yes. Answer choice. C. You look good. That graph pretty much matches up with what we drew and answer choice. D. No. That doesn't work. That is the wrong formula. The intercepts don't work. So yes. Answer choice. C. Well done. We'll catch you on the next one.

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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)², x ≤ 1