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 of f and ƒ¯¹. ƒ(x) = x² − 4, x ≥ 0

Accepted Answer

Welcome back. I am so glad you're here. We are given this function F of X which is equal to x squared minus five halves and we're told that x is greater than or equal to zero. We need to do several things with this function. We need to find the equation for its inverse which will be f inverse of X equals something. We'll figure that out shortly. We need to graph both this function and its inverse draw a rough sketch of a coordinate plane. We have a horizontal X axis of vertical Y axis and they come together at the origin in the middle and we need to provide the domain and range an interval notation for both. The function and its inverse lots to do. I am going to start off by writing the function again, F of X equals x squared minus. I'm going to write this five halves as 2. because that's how the answer choices have it in decimals. And now we're going to work on the equation for the inverse of this function. So instead of F of x, I'm going to write why? So we have Y equals x squared minus 2.5. And now wherever there's a Y I am going to substitute an X. So instead of Y equals its X equals. And wherever there's an X, I'm going to substitute a Y. So instead of x squared y squared Still -2. and now the goal here is to get that why completely by itself, why is miserable until it is alone? So let's add 2.5 to both sides of the equation. And on the left we have X plus 2.5 that equals Y squared on the right. And now to get why happy we will take the square root of both sides of the equation. So on the left we have plus or minus because we're taking the square root the square root of X-plus 2.5. And on the right, that is equal to why. And now through this process, we have got our equation for the inverse of our function. It is that plus or minus the square root of X plus 2.5. Great. Well, we almost have our equation. Note this plus or minus. We're going to have to figure out if it's plus or minus shortly. But we need a little bit more information about our domain and range in order to do that. So let's go back to our original function and we can graph that in order to graph it, I'm going to create an xy table figure out some values for X, plug those in to the original function and then figure out the corresponding Y values and plot those points. Note however, that we have a domain restriction for our X X must be greater than or equal to zero. So when choosing our X values, we are restricted by that. I'm going to choose 01 and two for the X values. Now plugging those into the original equation instead of x squared, it's going to be zero squared minus and again I'm going to use five point excuse me 2.5 not 5.2 I'm going to use 2.545 halves. So zero squared minus 2.50 squared is zero so zero minus 2.5. Well that equals a negative 2.5. Now plugging one in we have one squared minus 2.51 squared is one so one minus 2. that equals a negative 1.5. Now, plugging two in two squared minus 2. squared is four. So four minus 2.5 equals a positive 1.5. There are three points. Now let's plot them starting at the Origin 00 in the middle for the 1st 0.0 negative 2.5. We're going to go neither to the left nor to the right for our X value of zero and we're going down 2.5 units for our Y value of negative 2.5. I'm going to label this one because this is our Y intercept zero negative 2.5. Next 0.1 negative 1.5 starting from the Origin will go to the right one unit for our X value of one and will go down 1.5 units for our Y value of negative 1.5. So approximately there now for our third point to 1.5. Starting at the origin will go to the right two units for X value of two and we'll go up 1.5 units for our Y value of 1.5. So that's approximately here. Alright now to graph this, we start at zero, negative 2.5 and then this will curve upward toward positive infinity for both the X and Y axis. We can label this our F of X. Now let's write down what our domain and range are. Our domain is all of the possible X values for this function. So our X values as we were given are going to start at zero. So we'll put a bracket to say inclusive of zero and then we put zero comma infinity and then we put a parentheses for infinity. And that's our domain. Now, for the range that's all of the possible y values for this function. While our range starts at negative 2.5, inclusive of negative 2.5. And then heading up to positive infinity. So there is our domain and range for this original function. Now, let's take a look at this in verse. Okay, again, typically we set up an XY table but we want to keep in mind that we have domain and range restrictions here. Our domain for the inverse function is going to come from the range of our original function. The domain of our inverse is the range of the original. So here the domain is from inclusive of negative 2.5 up to positive infinity. And very similarly, the range of our inverse function is the domain of our original function. It is going to be inclusive of from zero up to infinity. And so now choosing our X values to plug into our table and then plug into our original function. We need to keep in mind this domain restriction, we need to start from negative 2.5 and head up to infinity. And then our why values need to be within this range from zero to infinity. So for our X values, we recall from previous lessons that in verses are just flipping the X and Y values from our original functions. So I'm just going to take the Y values that we got in our original function negative 2.5, negative 1.5 and positive 1.5. I'm going to use those and we should get these X values as the corresponding values for each, each one that we're putting in and those will be our new Y values. Keep in mind that we have two possible equations that we're trying to test out for our inverse equation. We have a plus square root of X plus 2.5 and we have a minus square root of X plus 2.5. We need to figure out which of those gives us answers within the the possible range here. So let's start plugging in. We're going to put in a square root of negative 2.5 plus 2.5. And inside of that Radical negative 2.5 plus 2.5. Well that equals zero. So we've got the square root of zero and that equals zero. So we have a positive zero and a negative zero. Well those are just both zero. So either equation works for that X value. Now let's plug in negative 1.5. So we have the square root of negative 1.5 plus 2.5. And inside the radical negative 1.5 plus 2.5 is one. So we have the squared of one which is one. So we have a positive one and a negative one. Which one fits our range. Well only the positive one fits the range that is from zero to infinity. The negative one needs to be rejected. So we have associated value of one. Which is correct given our original function. Now let's plug in 1.5. We have the square root of 1.5 plus 2.5. Well inside the radical 1.5 plus 2.5 equals four. So we have the square root of four that equals two but we have a positive two and negative two. Which one is within our range. Only the positive to the negative two is rejected. So we can reject this minus square root of X plus 2.5. And so our answer is only plus square root of X plus 2.5. And we don't usually write the plus in front of the square root. So there is our inverse functions equation, the square root of X plus 2.5. We rejected the negative square root of X plus 2. possibility. And here I'll fill this in. Now we've got three points and now we can graph these. So starting on our coordinate plane at the origin 00 for our first point we have negative 2.5 and zero. So for our X value we're going to go to the left negative 2.5 and we're going neither up nor down for our Y value of zero. So that will be about here and I will label this because this is our X intercept for this inverse function. Now for our next point negative 1.51 from the origin. We're going to the left 1.5 units for X value. And we're going up one unit for our Y value. So approximately there now for our last 0.1 point 52 from the origin, we're going to the right 1.5 units for our X value and we're going up two units for our Y value. So approximately here. Now this starts at negative 2.50 and then it curves this way toward positive infinity. We can label this R f inverse of X. And looking at this even though it's very rough, it does show symmetry about the line Y equals X. So this is our inverse function. Alright? We've done it. We found the equation for the inverse. We graphed both of them and we have figured out their domains and ranges. This is wonderful. Now we just need to figure out the correct answer choice. So when we're looking at are possible answer choices. Let's look for these intercepts for the F of X. We'll look for a Y intercept at zero negative 2.5. And for our f inverse of X. Well, we'll look for an X intercept at negative 2.50. So starting with answer choice A F of X has that one has an X intercept at negative 2.50. That's not going to work. Sorry, answer choice A. You're rejected. Now, let's look at answer choice. B. Answer choice B has F of X with a Y intercept at zero, negative 2.5. All right. It has F inverse of X with an X intercept at negative 2.50. Yes, that's great. Let's check out the other information. There is an F inverse of X equals the square root of X plus 2.5. That is our equation that we came up with the domain of F of X equals zero to infinity. Yes. That's what we came up with. The range of F of X is from negative 2.5 to infinity. Yes, That's what we have domain of F inverse of X, from negative 2.5 to infinity. Yes. And the range of F inverse of access from zero to infinity. Yes, the graph looks pretty darn similar. This is it? All right, Answer? Choice B. We choose you. Well done. We'll catch 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 of f and ƒ¯¹. ƒ(x) = x² − 4, x ≥ 0

Key Concepts

Inverse Functions

Domain and Range

Graphing Functions

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