Algebra 2 Inverse Functions and Relations

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Today we will be covering inverse functions and relations.0002

And this is a topic (inverse relations) that we introduced a little bit in Algebra I.0006

But we will be going into much greater depth today.0011

First, reviewing the definition of the inverse of a relation: recall that a relation is a set of ordered pairs.0014

The inverse relation of relation R (and the notation is this, R^-1, or if you had a different letter,0022

when we are talking about functions, then you would put that letter, and then -1,0033

to express that you are talking about the inverse of a relation) is the set of ordered pairs0037

in which the coordinates of each ordered pair of R are reversed.0047

So, let's think about what this means.0051

First of all, a relation is a set of ordered pairs; and that would be something like {(2,4),(3,-7),(2,6),(8,-4)}.0056

And this is a relation, because there is a correspondence between the first element and the second element.0070

You could write these out as a table; and each of the first elements has a correspondence with a member of the second set.0078

So, this is the relation; the inverse of R, R^-1, would be when the first and second coordinates are switched.0088

What we are doing, then, is interchanging the domain and the range.0101

So, 2, 3, 2, and 8 are members of the domain; and 4, -7, 6, -4 are members of the range; now it becomes the opposite.0105

Now, 4, -7, 6, -4 are the domain, and these other values are the range.0127

I also want to point something out: let's write R as a table, and I have my x-values, 2, 3, 2, 8, and my y-values, 4, -7, 6, -4.0134

This is a relation, because there is a correspondence between members of the x (the domain), and members of the range (the y-values).0149

2 corresponds to 4, 3 to -7, and so on.0158

So, this is a relation; however, it is not a function; and that is because you see that 2 is assigned to two values of the range.0162

And in a function, each member of the domain is assigned to only one member of the range.0175

So, this is a relation--yes; function--no; it is not a function.0183

Functions are relations, but not all relations are functions; functions are a subset of relations.0196

Let's talk more about functions right now.0201

The inverse function (f^-1(x)) is a special case of an inverse relation; the inverse of a function f(x) is a special case.0204

And, as I talked about, there are some restrictions on functions.0216

Functions must meet the criteria that each element of the domain is assigned to only one member of the range.0222

f^-1(b) = a if and only if f(a) = b.0232

The domain and range of the inverse function are the range and domain of the function.0240

Just as I talked about with relations, the domain and range are flipped; they are switched around0247

when you go from the function to the inverse, or from the inverse to the function.0255

And let's think about what this is saying right here: f^-1(b) = a if and only if f(a) = b.0260

Let's look at what that would mean in a general case.0269

If I have some function, and then I said part of this function is the ordered pair (3,4),0276

if it has an inverse, then that means that the values are going to be reversed.0285

If I looked for f(3), I would get 4 out; if I went ahead and put 4 into the inverse, my result would be 3; domain and range are switched around.0296

Taking a more specific example, if f(x) = 2x - 1, and f^-1(x) is (x + 1)/2, let's say I was asked to find f(6).0310

Then, I would substitute that 6 in here to get 11.0331

My expectation would be that f^-1(11) would be 6, because, if f(a) = b,0338

and here we are letting a equal 6 and b equal 11, then f^-1(b), f^-1(11), should equal 6.0349

So, let's check out if that holds up--let's see if it does.0362

I am asked to find f^-1(11), which would be 11 + 1, over 2, equals 12/2, equals 6.0364

So, f^-1(11) is 6, and f(6) is 11; so, this holds up and supports the fact that these two are inverse functions of each other.0377

There is a certain set of rules that you can follow to actually construct an inverse function.0402

So, if f(x) is given, f^-1 can be constructed by following this procedure.0407

First, write y instead of f(x): change the f(x) notation to y.0412

Then, interchange x and y: wherever there is an x, make it a y; wherever there is a y, change that to an x.0418

The hardest step is this one: you need to then solve for y, which, depending on the function, can be somewhat complicated.0427

Then, write f^-1(x) for y: once you have solved for y (you have isolated y on the left side of the equation),0434

you rewrite that as f^-1, because that is what you are looking for.0442

For example, f(x) = 2x + 7; if you wanted to find f^-1(x), step 1 is to write y instead of f(x).0446

1, 2, 3, 4 steps: the first step is to write y instead of f(x).0459

The second step is to interchange x and y: I am going to change that to an x; I am going to change that to a y.0468

The third step is to solve for y: I am going to subtract 7 from both sides--I am trying to isolate y.0478

Now, I need to divide both sides by 2.0487

I have y isolated, but I am just going to rewrite this in a more standard form, with the isolated variable on the left side of the equation.0495

So, I have y = (x - 7)/2, and now I am going to write it as f^-1(x), instead of using a y here.0502

I have found the inverse of f(x) by following these four steps.0518

One more example of constructing an inverse function: given f(x) = 5x - 3, we are going to find f^-1(x).0528

Step 1: Change f(x) to y--let's rewrite this over here and follow each step down.0542

So, step 1--I am going to change this to y = 5x - 3.0557

Step 2: Interchange x and y--I am going to make this an x; I am going to make this a y.0562

Then, solve for y.0579

This was step 1; step 2; solving for y--I am going to add 3 to both sides;0586

then I need to divide both sides by 5; so I get (x + 3)/5 = y; and I am just going to rewrite that with the y on the left, in standard form here.0595

OK, next, replace y with f^-1(x)--just change the notation.0609

This is going to be f^-1(x) = (x + 3)/5.0626

So, I found the inverse of this function by using this four-step procedure.0635

Inverses and compositions: in the last lesson, we talked about composition functions.0644

And now, we are going to discuss how this relates to inverses, and how it can help you to determine if two functions are inverses of each other.0652

The functions f and g are inverses if and only if f composed with g of x equals x, and g composed with f of x equals x, for all x.0663

So, if I put a number in here (2), and I evaluate it for the composite function, I will get that same value back.0678

If I use 2, I will get 2 back; if I use 50, I will get 50 back; and the same when I look at g composed with f.0691

If I evaluate that for a particular value, 7, I will get 7 back.0699

Let's go ahead and take an example to illustrate this.0707

Given f(x) = 6x - 1, and g(x) = (x + 1)/6, are they inverses?0710

I can use this to evaluate whether or not two functions are inverses of each other.0725

And if they are inverses, then when I take f composed with g of x, I should get x back.0732

Let's look at what f composed with g of x is: recall that is the same as saying f(g(x)).0743

Replace this with g(x): well, g(x) is (x + 1)/6; f of this can be found by looking at the function and replacing x with this algebraic expression, (x + 1)/6.0753

So, f((x + 1)/6) is 6 times (x + 1)/6, minus 1.0771

And here, these 6's end up cancelling out; so that leaves me with x + 1 - 1 = x.0788

So, f composed with g of x does equal x; that is half of what I have to figure out.0802

Now, does g composed with f of x equal x?0810

Well, g composed with f of x equals g(f(x)); well, that is g(6x - 1), so I need to evaluate g for this algebraic expression.0816

Well, g is (x + 1)/6, so I need to substitute 6x - 1 right here, and then I need to add 1 to that value and divide it by 6.0835

I am going to remove these parentheses to get (6x - 1 + 1)/6.0850

The 1's cancel each other out to leave 6x/6; the 6's cancel to get x; so, g composed with f of x equals x.0857

And f composed with g of x is x, so are they inverses? Yes, f(x) and g(x) are inverses of each other.0871

So, this fact right here allows us to determine if two functions are inverses of each other or not.0879

OK, the first example: find the inverses of the relation (I am going to call this relation R, and I am asked to find the inverse of R).0890

Remember that, in the inverse, I flip around the two values; these domain values and the range values are going to be reversed.0901

That is going to give me {(3,1),(4,2),(7,3),(8,1),(9,4),(6,3)}; and then, I am just double-checking that they are all correctly reversed; and they are.0913

This is the inverse of the relation R.0934

Next, find the inverse of g(x) = 2x + 4.0940

All right, the first step is to change g(x) to y; that is going to give me y = 2x + 4.0946

Second, interchange x and y: OK, this is going to become x; this will become y.0959

The third step: solve for y--I am going to subtract both sides by 4: x - 4 = 2y.0973

Next, I am going to divide both sides by 2: (x - 4)/2 = y.0986

I have isolated y; I am just going to rewrite this with y on the left.0993

And then, I am going to replace y, or change y to the notation g^-1(x).0998

I found that the inverse of this function is (x - 4)/2 by following these four steps.1016

Example 3: Find the inverse of h(x) = 2/3x - 4.1027

Again, here it is h(x), so change h(x) to y; that is going to give me y = 2/3x - 4.1032

The second step: interchange x and y--this is going to give me x = 2/3y - 4.1045

Next, solve for y: I need to add 4 to both sides, and then, after than, multiply both sides by 3/2.1059

And that is going to give me 3/2(x + 4) = y; multiplying this out, that is going to give me 3/2x + 3/2(4) = y.1076

3/2x...and this is 3 times 4 is 12, divided by 2; that becomes 6.1094

Finally, I am going to change y to h^-1(x).1102

I am going to do that at the same time as I am going to go ahead and put the y on the left side, and this becomes h^-1(x) = 3/2x + 6.1111

So, the inverse of the function given is h^-1(x) = 3/2x + 6--again, I did this by following the four-step procedure.1122

Given f(x) and g(x), we are asked to determine if f and g are inverses of each other.1137

So, are f and g inverses of each other?1145

Now, recall that two functions are inverses of each other if and only if this scenario holds up.1148

If f composed with g of x equals x, and g composed with f of x equals x, then the functions are inverses.1156

OK, so let's go ahead and find f composed with g of x, which equals f(g(x)).1182

OK, f(g(x))--well, that is f of this expression--f(1/5(x + 8)).1194

Well, f(x) is 5x - 8, so f of this expression would be 5; then, replacing x with this, that is 1/5(x + 8) - 8.1207

OK, let's get rid of these parentheses and do some multiplying.1224

This is 5 times 1/5(x + 8) - 8; these 5's cancel, and I am just left with a 1 here; so that is just (x + 8) - 8 = x + 8 - 8 = x.1228

So, we have our first part: f composed with g of x is x; now, let's see if it works out for g composed with f.1250

Well, recall that that is going to be the same as g(f(x)).1262

Up here, f(x) is 5x - 8; so g(5x - 8) is what we are looking for; I need to evaluate g for this expression, this function, when x is 5x - 8.1270

So, that is 1/5...of (5x - 8); and then I am going to add 8 to that.1284

So, here I have 1/5...now let's get rid of some of these parentheses inside to simplify...5x - 8 + 8, getting rid of these...1312

That is 1/5, and then I have a negative 8 and a positive 8, so those become 0; so it is just 1/5 times 5x.1325

The 5's cancel out, so that equals x; so g composed with f of x is also x.1337

So, are these inverses of each other? Yes, f(x) and g(x) are inverses of each other, because these two hold true.1346

f composed with g of x is x, and g composed with f of x is also x.1364

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