Video: AQA GCSE Mathematics Higher Tier Pack 3 β€’ Paper 3 β€’ Question 27

Consider the functions 𝑓(π‘₯) = π‘₯ βˆ’ 2 and 𝑔(π‘₯) = 5 βˆ’ π‘₯Β². a) Show that 𝑔𝑓(π‘₯) = 1 + 4π‘₯ βˆ’ π‘₯Β². b) Solve 𝑔𝑓(π‘₯) = 𝑓𝑔(π‘₯).

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Video Transcript

Consider the functions 𝑓 of π‘₯ equals π‘₯ minus two and 𝑔 of π‘₯ equals five minus π‘₯ squared.

Part a) Show that 𝑔 of 𝑓 of π‘₯ equals one plus four π‘₯ minus π‘₯ squared.

𝑔 of 𝑓 of π‘₯ is a composite function. It means that we take an input value π‘₯, apply the function 𝑓 first, and then apply 𝑔 to the result. In this case though, we’re looking to find a general algebraic expression for the composite function 𝑔 of 𝑓 of π‘₯.

We start with an π‘₯-value. The function 𝑓 of π‘₯ is π‘₯ minus two, meaning we subtract two from our π‘₯-value. This then becomes the input for the second function, 𝑔. The function 𝑔 of π‘₯ is five minus π‘₯ squared. So we take our new input of π‘₯ minus two and replace π‘₯ with it, giving five minus π‘₯ minus two all squared.

We need to simplify this algebraic expression. So first, we’ll expand the brackets. π‘₯ minus two all squared means π‘₯ minus two multiplied by π‘₯ minus two. You can use the FOIL method to help with this expansion if you wish. π‘₯ multiplied by π‘₯ gives π‘₯ squared. π‘₯ multiplied by negative two gives negative two π‘₯. Negative two multiplied by π‘₯ gives another lot of negative two π‘₯. And negative two multiplied by negative two gives positive four.

We can then simplify this expansion by grouping like terms. In the centre, we have negative two π‘₯ minus two π‘₯, which is equal to negative four π‘₯. So we can replace π‘₯ minus two all squared with our expanded version of π‘₯ squared minus four π‘₯ plus four.

Now we need to expand this larger bracket which has a negative sign in front of it. This means that we’re multiplying the whole bracket by negative one, which just has the effect of changing all of the signs. So the five stays the same. We then have negative π‘₯ squared plus four π‘₯ minus four.

We’re nearly there with the simplification. But we just have one final step, which is to group the like terms. In this case, the only like terms are the five and negative four, which are both constant terms. They have no π‘₯s or π‘₯ squareds. Five minus four is one. And we’ll just swap the order of the other two terms around to be consistent with the way 𝑔 of 𝑓 of π‘₯ is written in the question. We’ve shown then that 𝑔 of 𝑓 of π‘₯ is equal to one plus four π‘₯ minus π‘₯ squared as required.

Part b) of the question says, β€œSolve 𝑔 of 𝑓 of π‘₯ equals 𝑓 of 𝑔 of π‘₯.”

Here the two functions have been composed in different orders. 𝑔 of 𝑓 of π‘₯, remember, means we apply 𝑓 first and then apply 𝑔, whereas 𝑓 of 𝑔 of π‘₯ means we apply 𝑔 first and then apply 𝑓. Applying functions in different orders does not give the same result in general.

We have an algebraic expression for 𝑔 of 𝑓 of π‘₯ from part a). It’s one plus four π‘₯ minus π‘₯ squared. But we don’t have one for 𝑓 of 𝑔 of π‘₯. So our first step is going to be to find this composite function. 𝑔 of π‘₯ is the function five minus π‘₯ squared, and then we’re applying 𝑓 to this. 𝑓 of π‘₯ is the function π‘₯ minus two, meaning we take our input value and subtract two.

So if our input value is five minus π‘₯ squared, then we now have five minus π‘₯ squared minus two. This simplifies to three minus π‘₯ squared. So this is our algebraic expression for 𝑓 of 𝑔 of π‘₯.

Next, we’ll take our two algebraic expressions for 𝑔 of 𝑓 of π‘₯ and 𝑓 of 𝑔 of π‘₯ and substitute them into the two sides of the equation. 𝑔 of 𝑓 of π‘₯ is one plus four π‘₯ minus π‘₯ squared, and 𝑓 of 𝑔 of π‘₯ is three minus π‘₯ squared. So we have the equation one plus four π‘₯ minus π‘₯ squared equals three minus π‘₯ squared.

Notice that both sides of this equation have a term of negative π‘₯ squared. So these will cancel each other out. You could also see this by adding π‘₯ squared to each side of the equation. Once these terms have been cancelled out, we’re left with the linear equation one plus four π‘₯ equals three, which we need to solve for π‘₯.

To do so, we first subtract one from each side, giving four π‘₯ equals two, and then divide both sides of the equation by four, giving π‘₯ equals two over four. But of course, the fraction two over four should be simplified to one over two or one-half by dividing both the numerator and denominator by two.

So we’ve solved the equation 𝑔 of 𝑓 of π‘₯ equals 𝑓 of 𝑔 of π‘₯ and found that π‘₯ is equal to a half. You could of course substitute this value of a half back into the two composite functions and confirm that they do indeed give the same answer.

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