Lesson video

Thank you for joining us for this "transformations of graphs" lesson.

My name is Ms. Davis, and I'm gonna be helping you as you work your way through.

There's lots of exciting things to explore today, as well as things to recap that we've looked at before.

So make sure that you're in a nice quiet place that you can explore this, and that you pause the video and have a real think each time I offer you a question.

And then I'll help you with any hints, any ideas, that will help you develop your mathematics skills.

I'm really excited to explore this topic.

I hope you are too.

So let's get started.

Welcome to this lesson on "checking and securing understanding of moving between function notation and the definition".

Today you'll be able to fluently move between function notation and the function it represents, including when the function has been altered, and we're gonna have a bit of fun playing around with these functions today.

So a function is a mathematical relationship that uniquely maps values of one set to the values of another set.

We're gonna start then by looking at simple manipulation of functions.

We can use function notation to efficiently write and manipulate functions.

A quick recap, this notation means the function of x.

It is read as "f of x".

If we define the function that f of x is representing, then instead of writing the whole function, we can use the notation, and that's gonna save us time.

For example, if I have the function f of x equals 5x minus four, we can evaluate this function for different values of x.

F of one, for example, would be one.

f of two would be six, f of three would be 11.

And we could put these into a table of values if we wished.

So there's a table of values, I've chosen some inputs, and then I've worked out the outputs that they map to.

Of course, we could also use that to graph the function if we wished.

Function notation could be especially useful when we wish to manipulate a function.

So we can evaluate the below for any valid value of x.

So here we've got f of x, and that means we can evaluate the function for the given value of x, like we did previously.

Now here we've got three f of x.

Now that means we can evaluate the function for the given value of x, and then multiply the result by three.

Three lots of f of x.

Now this one is different.

This one is read "f of 3x", and that means we multiply the given value of x by three first, and then substitute it into the function.

And here we've got f of x subtract two, which means we need to subtract two from the given value of x, and then substitute that value into the function.

Let's have a look at how that works.

So we're gonna evaluate each of these when x equals two.

So first, f of x.

Well f of two is five lots of two minus four, which is six.

So three f of x would be three lots of f of two, in this example.

So we can do three lots of five times two minus four.

We already did that in the previous question.

That's three lots of six, which is 18.

F of 3x.

Well that means we do f of three times two, in this example, so that's gonna be f of six.

So we've multiplied the value of x by three first, and then we're doing the function of that value.

So f of six now is five lots of six minus four, which is 26.

This one here, f of x plus two, we didn't look at on the previous slide, so we'll do that now.

So that means we're gonna do f of two, and then add two.

So f of two is five lots of two minus four, and then we're gonna add two to that.

So that gives us eight.

And finally, when we've got x minus two in the bracket, that means we need to do the value of x minus two first and then input it into the function.

So two minus two is zero, so we're doing f of zero.

There's five lots of zero minus four, or negative four.

If you want to pause the video and just go through those again, feel free to do that.

So Sam is evaluating g of 2x when x is five.

The function g of x is 3x plus two.

They say g of five is 17, so I can double that to get g of 2x equals 34.

What mistake have they made? This may take you a little bit of thinking, so pause the video.

What do you reckon? Alright, well what they've done is they've evaluated two lots of g of x, because they've done g of five is 17 and then doubled it.

What they were trying to evaluate was g of 2x, so they'd need to multiply x by two first, then input into the function.

And that's a really important distinction.

So Sam says this is actually the same as evaluating g of 10.

Well yes, when x is five, two lots of five is 10, so they need to evaluate g of 10.

This is the important point.

When the function manipulation is inside the bracket, that change needs to be applied to x before inputting it into the function.

Quick check, then.

I'd like you to match the below function notation to the correct meaning.

Off you go.

Let's have a look.

F of x plus five means evaluate the function for the given value of x, then add five.

F of x minus five means subtracting five from the given value of x, then substituting this into the function.

F of 5x means multiply the given value of x by five, then substitute that into the function.

F of x plus five means add five to the given value of x, and then substitute that into the function.

Then five f of x means evaluate the function for the given value of x, then multiply by five.

We're gonna do more practise of this through the lesson.

Right, I'd now like you to evaluate each of these when x is three.

There's quite a lot to do, so take your time.

Off you go.

Let's have a look.

So we're doing g of three, then plus one.

Which is two lots of three plus one, plus one, which gives you eight.

Then we're gonna do four lots of g of three.

So that's four lots of two lots of three plus one, which is 28.

Then we're doing g of two lots of three, which is g of six.

So two lots of six plus one is 13.

And then we're doing g of three minus two, which is g of one.

Two lots of one plus one, which is three.

Again, pause the video if you need to look through those workings further.

When we want to evaluate a manipulated function for multiple inputs, it can be useful to write an expression for the required function.

So here's g of x.

How could we write an expression for g of x plus three? What do you think? Right, well we need to do the function g of x, and then plus three.

So it's 2x plus one, plus three.

Which is 2x plus four.

Alright, how is this different to g of x plus three? Let's look at this together.

So this time, we'd need to add three to x before inputting into the function.

What that means is we can replace every x with x plus three.

So here, g of x plus three will be two lots of x plus three, plus one.

We've replaced the x with an x plus three, 'cause that's our new input.

Now I can expand the brackets and simplify.

You get 2x plus six plus one, or 2x plus seven.

You can see that those two functions are different.

Right, what do you think we can do to write an expression for three g of x and g of 3x? Alright, well three g of x means three lots of the function g of x, so we need to do three lots of 2x plus one, which is 6x plus three.

Now g of 3x means we need to replace x with 3x.

So there's two lots of 3x, plus one, or 6x plus one.

Now we can check these by inputting a value.

So let's try x is four.

So if we do what we did previously, and we do g of four plus three, so we need to apply the function g of x and then add three, and that gets us 12.

If we use our new function, which we said was gonna be 2x plus four, let's try that.

Two lots of four plus four, and that is 12.

So g of x plus three is 2x plus four.

Let's try it for our next function, g of x plus three.

So g of four plus three would be g of seven.

So two lots of seven plus one, which is 15.

What happens when we substitute four into 2x plus seven? With two lots of four plus seven, that is 15, so that looks good as well.

Do the same with three g of four, which would be 27.

We look at 6x plus three.

That also gets us 27.

And g of 3x, well that would be g of 12.

Three lots of four, which is 25.

So let's try putting four into 6x plus one.

Also 25.

So we've just checked that our expressions are correct for a value of x.

Your turn! If f of x equals 4x plus three, which of these is an expression for f of x minus two? You might wanna write some steps of working to help you with this one.

So it is C.

Let's look at why.

So if we've got f of x minus two, we need to replace x with x minus two.

That's four lots of x minus two, plus three, which is 4x minus eight plus three, or 4x minus five.

Now this works exactly the same if you have multiple x terms in your expression.

So we've got f of x is x squared minus 2x.

I'd like you to fill in the blanks for these two methods to evaluate f of 5x.

So the first method is inputting x as two straight away.

The second method is writing an expression, and then substituting x as two.

See if you can get those correct.

Let's have a look.

So f of five times two would be f of 10.

So we'd have 10 squared minus two lots of 10.

Which is 100 minus 20, or 80.

For our second method, if we were working out an actual expression for f of 5x, we do 5x squared minus two lots of 5x, which would be 25x squared minus 10x.

Now we've got that expression, we can substitute x equals two.

So 25 lots of two squared, minus 10 lots of two, which is 100 minus 20, or 80.

So we can see that our expression was correct, because when we evaluate that for x equals two, you get the same as when we did it the other way.

Now for this example, that first method was definitely quicker.

It'll depend on the expression that you are using.

It'll also depend on whether you want to input for multiple values of x.

Time to put that into practise.

You've got a function of x.

I'd like you to evaluate each of these when x is four.

Give it a go.

For two, you've got a function g of x.

I'd like you to write the simplified expression for each of these functions.

Off you go.

Same idea for question three, but I picked a slightly trickier function.

You've got g of x is x-squared plus 3x.

Off you go.

Question four, you've got h of x.

I'd like you to evaluate h of 3x when x equals negative four.

Then write an expression for h of 3x.

And then use your expression to check your answer to A.

Also, you can use your expression to evaluate h of 3x when h is five.

Give those a go.

Let's have a look.

So f of four is gonna give you 20.

2f of four will be 40.

F of 5x.

When you do f of five lots of four, which is f of 20, which is 68.

For D, f of x plus two.

So that's f of four plus two, which is f of six, which is 26.

Now f of x, then subtract one, is gonna be f of four subtract one, which is 20 subtract one, or 19.

Then f of negative x, where negative x is negative one times x.

So if we're doing negative one times four, that's negative four.

F of negative four is three lots of negative four plus eight, which is negative four.

Pause the video if you need to look through those working.

So for two we've got g of 5x is four lots of 5x minus one, or 20x minus one.

For B, five lots of g of x.

So five lots of 4x minus one, which is 20x minus five.

For C, g of x plus four, so we need to replace x with x plus four.

With four lots of x plus four minus one, which is 4x plus 16 minus one, which is 4x plus 15.

G of x, all add four, is gonna be 4x minus one add four, or 4x plus three.

E, g of x minus two.

We've got four lots of x minus two minus one, which is 4x minus nine.

And finally, g of a half x.

So we just need to replace x with a half x.

So four lots of a half x minus one, which is 2x minus one.

For question three, so g of 2x, we just need to replace every x with 2x.

So 2x squared, plus three lots of 2x.

4x squared plus 6x.

For B, three lots of g of x, so three lots of x squared plus 3x, which is 3x squared plus 9x.

For C, we need to replace every x with x plus one.

So we've got x plus one squared, plus three lots of x plus one.

Bit more work here, 'cause we've got the product of two binomials, so x plus one times x plus one is x-squared plus x plus x plus one, or x-squared plus 2x plus one.

Then we need to add three lots of x plus one.

Plus 3x plus three.

Simplifying x-squared plus 5x plus four.

And finally, nice easy one, g of x plus three is just gonna be x-squared plus 3x plus three.

Hardest one was definitely C there, where we needed to square a binomial.

For question four, we're evaluating when x is negative four.

So we've got h of three times negative four, which is h of negative 12.

Substituting that, we get 29.

Be careful with your negatives there.

An expression for h of 3x will be five subtract two lots of 3x, or five subtract 6x.

So now we can substitute negative four into that and see what we get, and we do get 29.

And finally, we can now use that expression to evaluate h of 3x for any value of x we like, so let's try it when x is five.

Five minus six lots of five is negative 25.

Right, you've learned all the skills that we need.

We're now just gonna apply some more complex manipulation.

So we can apply multiple operations to the same function.

So if we have f of x is 3x plus two, what do you think f of 2x plus three would look like? So all we need to do is input 2x plus three instead of x.

So f of 2x plus three would be three lots of 2x plus three, plus two.

And we can expand and simplify.

You get 6x plus 11.

Let's check what this looks like when x is five.

So if we were doing f of 2x plus three, that'd be f of two lots of five plus three, which would be f of 13.

So if we do f of 13, three lots of 13 plus two is 41.

Now let's look at the expression we created, 6x plus 11.

So if we do six lots of five plus 11, we also get 41.

So you can see that expression is correct.

Alex says, "This is just like when we did compound functions.

If we are evaluating f of g of x, we could input into g of x first, then into f of x." Or your other option was to combine the functions first, work out what the new function f of g of x would be, and then input your value in.

It's the same idea.

You can either input the value first, or you can create an expression first, and then input your value.

Let's look at this one.

2f of x plus three.

So 2f of x means we evaluate the function of x, then multiply by two.

So 2f of x plus three means we evaluate the function for x plus three, then multiply it by two.

So it's just being careful with your priority of operations.

So you can see I've substituted x plus three instead of x, then I'm multiplying the whole thing by two.

Do a bit of expanding, simplifying, and then multiplying everything by two.

You get 6x plus 22.

How do you think 2f of x plus three will be different? Well, 2f of x means we evaluate the function for x, then multiply it by two.

So 2f of x plus three means we evaluate the function for x, multiply by two, then add three.

So let's have a look.

2f of x is two lots of 3x plus two.

Then we need to add three.

Which is 6x plus four add three, or 6x plus seven.

So we're following the normal rules for priority of operations.

So those are the three we've looked at so far.

Have a go at this one.

How could we write an expression for f of 2x plus three? Well, f of x plus three means we evaluate the function for x and add three.

So here we need to evaluate the function for 2x, then add three.

So we've got three lots of 2x plus two.

Add three.

Which is 6x plus two add three, or 6x plus five.

So there's the four different functions we've now created.

Your turn! If f of x equals 5x minus three, I'd like you to match the function notation to the correct function.

Again, there's several steps here.

You might wanna write down your working to help you out.

Let's have a look.

A matches with g, b matches with f, c matches with h, and d matches with e.

If you didn't get those correct, just spend some time looking back through your working and seeing where you've gone wrong.

So given that g of x is x-squared plus 2x, Alex wants to evaluate 3g of x plus one when x is five.

So lots going on here.

Alex says, "First I'm going to write an expression for three lots of g of x plus one." And theirs is working.

Read back through that.

Can you see what mistake he has made? Well done if you spotted that he didn't replace every x with the input x plus one.

If he's putting x plus one into this function, it should be x plus one, all squared, plus two lots of x plus one.

He is correct that you then need to multiply that by three.

"Thanks," says Alex, "I've corrected that now." So that is now the correct first step.

He says, "Now I can input five into the new expression." And we get 144.

Sam says, "It would've been easier to evaluate this without creating a new expression." Let's have a look.

So we need to do 3g of five plus one, which is 3g of six.

So if we evaluate g of six, then multiply it by three.

So six squared plus two lots of six, all multiplied by three.

Which gives us 144.

You're right, Sam! That was easier this time.

I think we can agree with Alex that it was a lot easier just substituting the value than creating a new expression.

Now, when evaluating for one value of x, it is often easier to do that.

However, having an expression for the function will be easier if we're inputting multiple values for x.

So we then wanted to evaluate 3g of x plus one when x is six or two or negative three.

Having the expression will save us time there, 'cause we just need to then input the new values of x.

Let's have a go.

Given that f of x is five minus 3x, I'd like you to fill in the blanks for these two methods.

Give that one a go.

So our first method is substituting five in first.

So we're gonna do f of two times five, all plus three, which is gonna be f of 10 plus three.

If we evaluate f of 10, we've got five minus three lots of 10, but then we need to add three.

Which is five minus 30, add three, or negative 22.

If we look at creating an expression first, we're gonna do five minus three lots of 2x, add three.

Which is five minus 6x add three.

Which is eight subtract 6x.

Now we can substitute x as five.

That gives us negative 22.

This time, they're both reasonably efficient.

The first method was probably still quicker, but because we have an easier function, creating the expression was not as tricky.

Time for you to have a practise.

I'd like to evaluate each of these when x is three.

Give those a go.

For question two, I'd like you to write an expression for each of these.

For question three, you've got g of x is 3x plus five.

I'd like you to evaluate 4g of x minus two, when x is five.

Then write an expression for 4g of x minus two, and use that to evaluate when x is one, two, and three.

Got the same for question four.

We've got a slightly trickier function this time, so take your time with the algebraic manipulation.

Off you go! Let's have a look.

Who got 5f of three plus seven? So five lots of two times three plus one plus seven, and that gets you 42.

For B, 4f of three minus one, which is 4f of two, which is four lots of two times two plus one, which is 20.

For C, we've got f of five times three plus two, which is f of 17.

So two lots of 17 plus one is 35.

For D, we've got f of two times three, all subtract five, so that's f of six, subtract five.

So you've got two lots of six plus one, subtract five, which is eight.

So you've got 3f of two times three, which is 3f of six, which is three lots of two times six plus one, which is 39.

And finally, you've got f of three plus one, all add four.

So that's f of four, add four.

So two times four plus one, add four, which is 13.

For question two, our expressions, I'd like you to pause video and check my working for these, it's really important where you've put the brackets.

It's a lot easier to read than it is to say aloud.

For A, you should get 8x plus six.

For B, 8x plus 20.

For C, 8x plus five.

For D, 8x plus three.

Then for E, we've got 16x plus four.

You see that we've doubled our input for x, and then we've multiplied that by four.

So we've got 16 lots of x plus four.

And then for F we've got 2x plus one.

Do make sure you check that working.

For three, so when x is five, we've got four lots of g of five minus two.

So four lots of g of three, so four lots of three times three plus five, which is 56.

As an expression, we get 12x minus four.

Again, take some time to look at those working if you didn't get that correct.

Now we've got that expression, we can just substitute different values of X.

So 12 lots of one minus four is eight, 12 lots of two minus four is 20, 12 lots of three minus four is 32.

You might also want to check your answer for A.

So if we do 12 lots of five minus four, you get 56, which was our answer for A.

Same for four.

So if we're doing h of 2x, all plus three, we're doing h of two times five, which is 10, add three.

So putting that into our function, 10 squared times two plus 10, and then add three, gets you 213.

Again, look at the working for how I've done my expression.

You need to make sure that you're substituting 2x for x everywhere it appears.

Expanding and simplifying, you get 8x squared plus 2x plus three.

Now we can use that to evaluate.

So when x is one, we get 13, when x is two, 39.

When x is three, 81.

Again, you might wanna check when x is five.

So you've got five squared multiplied by eight, plus two lots of five plus three, and you do get 213.

So it's a nice way to check that.

Fantastic! That was a really exciting lesson where we're pulling together lots of our algebraic manipulation skills to apply to functions.

If you carry on doing work with functions and transformations of graphs, you'll see where these skills come into play.

If you'd like to read through all the things that we've looked at today, pause the video and do that now.

Otherwise, thank you so much for joining us and I hope you choose to learn with us again.