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Hello, Mr. Robson here.

Welcome to Maths.

Great decision to join me today, especially seeing as we're problem solving with functions and proofs.

I love problem solving, you love problem solving, so let's solve some problems. Our learning outcome is I'll be able to use our knowledge of functions and proof to solve problems. Lots of keywords you're gonna hear throughout today's lesson.

Two of which are function and expression.

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

An expression contains one or more terms where each term is separated by an operator.

Two parts of our learning today, we're gonna begin by looking at problems with functions.

If f of x equals five x minus four and g of x equals x minus 11, we can write expressions for composite functions like g of f of x and f of g of x.

G of f of x would look like so, that's the function f of x going into the function g of x and simplifying of course.

F of g of x looks like so.

That's the function g of x being input into the function f of x giving us five x minus 59.

We can use the same skills to write expressions for all sorts of manipulations of the function f of x, such as two f of x, f of two x, f of x plus two and f of x plus two.

Let's have a closer look at what those four things mean.

Two f of x is two lots of the function f of x.

f of x plus two is two being added to the function f of x.

F of two X is when two x is input into our function.

So would make logical sense that f of x plus two is when x plus two is input into the function f of x.

So in the case of two lots of f of x, that's two lots of f of x or two lots of five x minus four.

We'll leave it like that rather than expanding those brackets.

That's a nice concise expression.

f of x plus two, well that's f of x and then we'll add two to that.

Let's simplify five x minus two.

F of two x input into two x into that function f of x would look like that.

Let's simplify, 10 x minus four.

f of x plus two.

Well, when we input x plus two into our function f of x, we get that.

Let's expand that bracket, add the like terms, simplify and we get five x plus six.

Quick check.

You can do that.

If f of x equals seven x plus 11 and g of x equals three lots of five minus x, can you write expressions for these six things? Pause and give it a go now.

Welcome back.

Let's see how we did.

Three lots of f of x would look like that.

That's seven x plus 11 being multiplied by three.

f of x minus 15.

Well that's our function f of x and then subtract 15.

Eight lots of g of x would look like that.

Again, you're welcome to leave that expression in its brackets or expand it out, either is fine.

g of x plus five, well that's our function g of x.

When we add five to it, we will expand that bracket and simplify those like terms for that one.

F of seven x, well that's seven x going into our function f of x giving us 49 x plus 11.

g of x plus eight.

That's x plus eight going into our function g of x which expands and simplifies to negative three x minus nine.

Andeep and Izzy are manipulating the function f of x equals x plus four.

Andeep says "I'm great at algebra.

I know that three times a times b equals b times a times three.

So I predict that three f of x equals F of three x." Izzy says "it's good to make predictions, but it's even better when we test them." Well done Izzy, good point.

Can you test Andeep's prediction? Is he right? Pause.

Have a play with this bit of mathematics.

I'll be back with a big reveal in a moment.

Welcome back.

Hopefully you noticed three lots of f of x gives us three lots of x plus four which expands to three x plus 12, whereas f of three x gives us three x plus four so it wasn't true.

Three f of x is not equal to f of three x.

Andeep says "I was wrong.

But I don't mind making errors in maths Izzy.

Every error is an opportunity to learn." Izzy says "I totally agree Andeep, I agree too.

Three f of x and f of three x are function notation so three f of x does not equal f of three x." But is that always true? Could there be a function f of x such that three f of x equals f of three x? Andeep says, "hold on Izzy, I found some cases whereby three f of x equals F of three x.

Can you think of any?" Is Andeep right? Is it ever possible that this is true? Pause, have a think.

See if you can come up with some examples.

Welcome back.

I wonder if you found any.

Andeep found some and he presents them to Izzy.

Izzy says "well done.

It's true for these two.

There will be some cases where three f of x equals F of three x." Quick check you've got there.

For a function f of x, it is true that a f of x equals F of axe where a is an integer.

Is that always true, sometimes true or never true? Pause and have a think.

Welcome back.

Hopefully you said sometimes.

In most cases a F of a will not equal F of axe, but there will be some cases where they are equal.

Sometimes we're given limited information and have to work backwards to find the expression of the original function or the manipulation that has happened to a function.

f of x equals kx minus seven.

Ooh, we're not given the actual f of x function here, but we are told F of three equals two and then challenged to find k where k is a constant.

Lots of ways we could do this.

If we input three into that f of x function, we know k multiplied by three minus seven will give us that output two because we're told F of three was equal to two.

There, we can set up this equation.

Three k minus seven equals two, so three k must equal nine and k must equal three.

Another example, g of x equals nine x plus one.

We have the expression of g of x, but we haven't got the manipulation in this case.

If kg of negative one equals negative 40, can you find k where k is a constant? Well, let's look at an expression for kg of x, that would be k lots of nine x plus one.

We know an input of negative one will give us an output of negative 40, so let's express that.

From there, we can do some simplifying and find that k multiplied by negative eight equals negative 40, so k must be equal to five.

Some examples like this might include composite functions.

We know f of x but we're not given all the information of g of x.

We're told it's kx minus four and then we're told that f of g two equals 56 and challenged to find k where k is a constant.

We could start with an expression for f of g of x, f of g of x will be the output of g of x kx minus four going into the function f of x.

We're gonna do some simplifying from there and we're happy at that point.

We're now gonna put our input two and reflect the fact that we know it gives us an output of 56.

Some simplifying and tidying from there and we find that 2k minus two equals eight, so 2k equals 10k equals five.

We can check that that's true.

We think that g of x is five x minus four, so let's find out if that's true.

G of two would therefore be five lots of two minus four, that's six.

When that goes into f of x, it should give us 56.

It does.

We know we're right.

We could have solved this same problem by considering inputs and outputs.

F of what makes 56? Well that what would be six.

Why do we care about that? Well, that's because six is what we need to input into the function f of x to get an output of 56 therefore g of two equals six.

G of two equals six, then k multiplied by two minus four equals six, k equals five.

Absolutely no surprise to arrive at the exact same answer.

In fact, it's reassuring if you can do the same problem in two different ways and you get the same answer, you are reassured that you are right.

Quick check.

You can do similar problems to those.

There's three problems here for you to think about and in each case we're told that k is a constant.

See if you can find the solutions to these three problems now.

Pause, see you in a moment for the solutions.

Welcome back.

For the first problem.

If f of three equals negative three, find k.

Let's reflect that input of three going into the expression and gives us an output of negative three.

If that's true, then three k plus 12 equals negative three.

Therefore k equals negative five.

For the second problem, an expression for kg of x helps.

We then reflect the fact that an input of nine gives us an output of negative 296.

Let's simplify that.

We find that k equals negative four.

For the final problem, g of x plus k.

We can express like so.

Let's reflect that input and output and we find that k equals 13.

Let's try a similar problem but with a composite function.

If g of f of five equals 311, can you find k where k is a constant in this example.

Pause, give this problem a go, see you in a moment so you can compare your work to mine.

Welcome back.

Lots of ways we could have done this.

One way is to ask ourself the question g of what makes 311? Well, g of what makes 311, that input needs to be 105.

Why do we care about that? Because f of five must equal 105 if g of f of five equals 311.

So if f of five equals 105, then 15k equals 105, k equals seven.

We can check that that's true.

We think f of x equals seven lots of x plus 10, so let's evaluate f of five and then input that output into the function g of x and we get to 311.

Our composite function mapped five to 311, so k must be equal to seven.

An alternative method was to write an expression for g of f of x.

Either method is perfectly valid.

Practise time now.

We're given two functions, f of x and g of x and we're asked to write expressions for these six things.

Pause and write those now.

For question two, we're given a function f of x and g of x and we're asked to show that f of x plus five minus two f of x equals 16 minus three x.

For part B, we're asked to show that f of four x is not equal to four f of x.

For part C, we're asked to show that g of x plus three minus g of x is equal to three lots of two x plus one.

Pause and see if you can prove those three facts now.

Question three, we're given the fact that f of x equals k, lots of X minus seven and g of x equals x squared plus two x minus 15.

for A, B and C, we're asked to find k where k is a constant and we're given an inputs and outputs for all three of those things.

Pause, see if you can solve these problems now.

Feedback time.

Let's see how we did.

Question one, part A, write an expression for eight f of x, we should have got 64 x minus 24.

For B, we should have got eight x plus five by the time you've simplified.

For C, we should've got four x plus 27.

For D, we should've got four lots of x plus six.

You might leave that expression bracketed.

You might expand those brackets.

Either is fine.

Part E, f of g of x plus five.

It's useful to write an expression for f of g of x first 32 x plus 221.

Now when we input x plus five into that expression, we get 32 x plus 381.

Part F was a little bit complicated.

We can undo some of that complication by writing an expression for g of f of x, which we four lots of eight x plus four.

Now we're gonna input two x and multiply the output by five.

That'll be reflected like so that simplifies to 320 x plus 80.

For question two, we are trying to show that a few things are true.

For part A, it helps to express f of x plus five and two f of x.

Once we've got those two expressions, we can show that f of x plus five minus two f of x does indeed equal 16 minus three x.

Part B, f of four x is 12 x minus one, whereas four f of x is 12 x minus four and they are not equal.

Part C, lets write an expression for g of x plus three.

We're gonna expand and simplify there and get the expression x squared plus four x plus eight.

When we take that and we subtract g of x, we end up with that which we simplify to six x plus three which factorises to three lots of two x plus one.

Question three, we are finding k, the constant in each example.

For part A, we'll input 10 and show that gives us an output of negative 21.

Therefore k must be equal to negative seven.

For part B, we can input seven express the fact that it gives us an output of negative 96 has a lot of simplifying scoring in that bracket there.

Once we've done that, we find that 48k equals negative 96, k is equal to negative two.

For part C, an expression for h of j of x would be three lots of kx squared minus four.

Let's show the fact that an input of five gives an output of 221.

That takes us to the point that 75k equals 225, so k must equal three.

Onto the second part of our learning now, problems with proof.

Sometimes in maths we come across proofs, which we know can't be right.

Sofia says, "did you know that two equals one Lucas?" And Lucas says, "no, it doesn't.

There's no way that's true." Sofia says 'it is true.

I can prove it." Well, Sofia, I'm looking forward to this.

Sofia writes, if x equals y, then x squared equals x multiplied by y.

Okay, add negative y squared to both sides.

Factorise, I'm with you.

Divide both sides by x minus y.

Okay, and then x is equal to y, so we can rewrite that.

Therefore, two y equals y divide both sides of the equation by y and two is equal to one.

Sofia, I'm stunned.

Two does equal one.

Sofia says, "I told you, I'm a genius.

I have broken maths." Or has she? Can you see any fault in Sofia's proof or is she right? Pause and have a good look at that work there.

Welcome back.

I wonder if you spotted it.

If we see a proof that we know cannot be true, we need to investigate.

In fact, there are some proofs in maths that are still being checked by the finest mathematicians on the planet.

We're gonna check this one because we know that two doesn't equal one.

The problem with this proof comes here in this step.

Can you see why? Well spotted if you can spot this.

The step is to divide both sides by x minus y.

At first glance, this looks fine, but when you remember that x equals y, the problem becomes clearer.

If x equals y, then x minus y equals zero and dividing by zero is undefined.

That means this is not a valid step in the proof.

Substituting in a value makes this error more obvious.

Rather than starting with x equals y, let's start with five equals five.

If five equals five, then five squared equals five times five.

It does.

Five squared minus five squared equals five times five minus five squared.

That's true, and we can factorised those and that remains true, but here when we've divided through by five minus five on both sides, we're now at a position of having a false statement.

Five plus five is not equal to five, so we can't say that two lots of five equals five and we can't say that two equals one.

Quick check that you can spot the error in proof.

Spot the error in this proof.

This proof proves that three equals two.

Alarm bells should be ringing.

Something must be wrong somewhere.

Pause and see if you can spot it.

Welcome back.

Did you spot it? At first glance, it appears perfectly valid, but if you solve this equation, you'd find that x equals three, therefore two x minus six equals zero.

Dividing by zero is not a valid step.

In order for it to be a proof, all steps need to be valid.

Sometimes an invalid step can convince us to believe something is true when it is not.

26 over 65 cancels down to two fifths.

It does.

Here's a proof.

In 26, we'll just scratch out the ones digit and in 65 we'll scratch out the tens digit.

That's why it's equal to two over five.

Here's another example.

16 over 64 equals a quarter.

Here's my proof.

Let's just scratch out the ones digit in the numerator.

The tens digit in the denominator, 16 over 64 equals a quarter.

Here's another proof.

I'm just gonna scratch out the ones digit in the numerator.

The tens digit in the denominator, 19 over 95 cancels down to one fifth.

What I'd like you to do is come up with a counterexample to show that this is an invalid step.

I can't just do this.

Scratch out the ones in the numerator, the tens in the denominator.

Can you come up with a counterexample to show that that is not a valid proof? Welcome back.

Lots of examples you could have used here.

You might have shown 12 over 24 when you scratch out the ones digit in the 12 and the tens digit in the 24, we show that 12 over 24 is equal to one quarter, which is not an equivalent fraction.

That's not true.

This is not a valid proof.

Many other counterexamples were acceptable.

In order for it to be a proof, all steps need to be valid.

This was a case of some very invalid steps.

This proof is also clearly wrong.

We're proving that pi equals three.

It doesn't, but can you see why it's wrong? Have a good look at that proof.

See if you can spot where it's gone wrong.

Welcome back.

Let's have a close look.

The problem with this proof comes here.

We have to be very careful when taking the square root of both sides of an equation.

X squared equals 25 does indeed have two solutions.

X equals positive five and x equals negative five, but if A squared equals B squared, we might say that A equals positive or negative B, but it's not necessarily the case that both are true.

For example, if A equals two and B equals negative two.

It is true that A squared equals B squared, but when we work backwards, it's not the case that A equals positive B or negative B.

This proof assumes that the positive root is correct, but we can see from the result that this was not the case because pi is not equal to three.

If we take the negative root at this step, we get back to the start, three plus pi equals two x.

We have to be very careful when taking the square root of both sides of an equation.

Sometimes we get two truths, sometimes we get one truth and one falsehood, so be careful.

It's not just in algebra that we may see suspicious proofs.

When these four shapes are placed together, they form an eight by eight square with an area of 64 square units.

When we arrange the four pieces to make a rectangle, they form a 13 by five rectangle with an area of 65 square units.

My key question for you, where has the extra square come from? Pause and see if you can spot it.

Welcome back.

I wonder if you noticed.

Your eyes have been deceived by what we call a dissection fallacy.

The problem is that the triangles are not identical.

In the red triangle on the left hand side of the screen, the gradient of the hypotenuse is three over eight.

This is a straight line.

When we look at the red triangle or is it a triangle? We see that it starts with a gradient of two over five and then changes to have a gradient of one over three.

This is not a straight line, so that shape there is in fact not a triangle.

It's a quadrilateral.

Your eyes have been deceived by a dissection fallacy.

If we insert congruent triangles, you can see the space that created the extra one, that little gap in there.

Quick check you've got that.

In geometrical proofs, we can be deceived by a bisection fallacy, dissection fallacy or dissecting falsely.

Pause and answer this one.

I hope you said B, dissection fallacy.

To avoid falling for a false proof, pay careful attention to the detail in every shape in the dissection.

Practise time now.

Question one, A equals zero, B equals two and C equals four and there's a proof.

A proof that shows that four is equal to zero.

Alarm bells are ringing, something wrong In that proof.

What I'd like you to do is find the error in the proof and write a sentence explaining why it's an invalid step.

Pause and do that now.

Question two.

If A equals B equals C, then look at that proof, three is equal to one.

Well, of course it isn't, which means there's an error somewhere.

I'd like you to find the error in this proof and write a sentence explaining why it's an invalid step.

Pause and do that now.

Question three.

I'd like you to write your own proof to show that zero equals 100.

My hint for you is you could start with A equals 50, B equals 100 and C equals zero.

I enjoyed doing this problem myself.

I hope you will too.

Pause and give it a go.

It might take you more than one attempt.

Question four, I'd like you to explain why in the second diagram there's an extra square.

Pause and have a good study of this screen.

Feedback time now, finding the error in this proof and writing a sentence explaining why it's an invalid step.

The problem is here.

The positive root of each term was taken.

It is wrong to assume that the positive roots are always the correct ones.

Question two.

Again, finding the error and the error was here.

If B equals C, then B minus C equals zero.

Dividing by zero is not a valid step.

In order for it to be a proof, all steps need to be valid.

Question three, I challenged you to write your own proof to show that zero equals 100.

You might have written if A equals 50, B equals 100 and C equals zero, then C plus B equals two A and so on and so on and so on and so on and so on and so on and so on, and therefore zero equals 100.

That is an example of a proof of zero equaling 100, but we know of course that taking only the positive roots here has invalidated the proof.

However, we could use this to convince a less vigilant mathematician that 100 equals zero and wouldn't that be fun? Question four, I asked you to explain why in the second diagram there's an extra square.

We're being fooled by a dissection fallacy.

This is not a triangle.

It is a quadrilateral.

The gradient of the red triangle is three over eight, the orange triangle, two over five.

That is not a straight line.

When we reverse the triangles and put the one with the greater gradient first we create space for the extra square.

If we overlay the respective gradients from the second shape, you can see more clearly the extra space created.

Well, we're at the end of the lesson now sadly, but what have we learned? We've learned that larger functions and proof can be used to solve a wide variety of problems. If we know f of x, then we can write expressions for manipulation such as four f of x and f of four x and know that they are not necessarily the same.

Invalid steps in algebraic proofs can be identified and incorrect steps such as dividing by zero or making assumptions around the square roots of an equation can be spotted.

Hope you've enjoyed today's lesson as much as I've enjoyed it? And I look forward to seeing you again soon for more maths.

Goodbye for now.