A-Level Maths: B8-10 Functions: The Domain of a Composite Function

[@helenday5031]

This lesson included some handy tips generally for handling numbers, which I found very helpful, such as the bit where the tutor multiplied both the top and bottom of the fraction by x. Hopefully I will absorb such tips and remember them!

[@AceOfHearts001]

Do you have an extension video touching on the range of a composite fumction?

[@aunabbas6192]

you are an amazing teacher. Deliver your lecture very well!!!

[@HoneyBee-wr2hv]

YOUR A KING, THANKYOU SO MUCH FOR THIS

[@AledorLearns]

Your lectures are mindblowing...

[@kabelan9660]

hi, i was wondering if you could help with this question. It's given that f(x) = x^2-4 ,x ∈ ℝ,x>8 and g(x) = 2x-2,x ∈ ℝ, x>3. The question was to determine the domain and range of fg(x). I calculated fg(x) to be 4x^2-8x but i'm still not sure how to work out the domain and range

[@TLMaths]

Hi, so fg(x) = (2x - 2)^2 - 4 = 4x^2 - 8x The domain of fg(x) is the intersection of the domain of g(x) and the domain of fg(x). Now the domain of g(x) is x > 3, and the domain of fg(x) shows no apparent restriction, so at the moment we just have x > 3. However, as f(x) also has a restricted domain, we need to check if all of the values of g(x) can be substituted into f(x). If x = 4, for example, then it can be substituted into g(x) to get g(4) = 6. But then 6 cannot be substituted into f(x) as it has a domain of x>8. So g(x) > 8, 2x - 2 > 8, so x > 5. So the domain of fg(x) is x ∈ ℝ, x > 5. When x = 5, fg(5) = 60. fg(x) describes a parabola with a vertex at (1,-4). So the range of fg(x) is fg(x) > 60 So the blue part of the curve here is fg(x): www.desmos.com/calculator/oaccqawzse

[@princeandrew9073]

ABSOLUTE KING

[@larry8160]

Excellent video, but I'm a bit blurry on a few things. How would you go about drawing the graph of x+2 with the domain not including -2? And why was the domain for gf(x) restricted to -1/2?

[@TLMaths]

You would draw a line, but where the 'hole' is, you would draw a little open circle to identify that point is not included.

[@TheGamingWattsit]

How do you find the ranges of each composite function

[@TLMaths]

It would be the range of the function given the domain as found in the video. So the range of fg(x) for example would be all possible y values apart from 0.

[@aunabbas6192]

How to find the domain if the domain of f(x) is restricted lets say 3《 X《7. And they say find the domain of g(x) for which the composite function gf(x) exist?

[@TLMaths]

You would have to find the range of f(x) with this restricted domain. This would then become the domain of g(x).

[@aunabbas6192]

@@TLMaths oh. So basically for a composite gf(x) to exist the range of f(x) should satisfy the domain of g(x). Now that makes complete sense. Thankyou Sir!

[@isaacade2075]

please could you explain that

[@AnnaHazel17]

If you had three functions (I'm talking hypothetically, so they could be anything - I'll call them f(x), g(x) and h(x)) and you were working out the domain of the composite function fgh(x), would both the domains of g(x) AND h(x) apply?

[@TLMaths]

It would be the intersection of the greatest possible domains of h(x), gh(x) and fgh(x)

[@MathBuz]

For fg(x), if we input value of x = -2 1/(1/(-2+2)) = 1 / (1/0) = 1 / infinity = 0 it gives the same answer as we put x = -2 before and after simplification that is zero. THEN why x = -2 is not included in domain of composite function?

[@TLMaths]

You have to be VERY careful here, 1/0 is undefined, you cannot write that it is equal to infinity. Likewise, you cannot write 1/infinity. You must not write infinity into an equation, treating it like a number. So when x = -2 is substituted into g(x), it is undefined. i.e. x = -2 is not in the domain of g(x), and so it cannot be in the domain of fg(x).

[@MathBuz]

TLMaths But sometimes we use 1/infinity = 0 For example when we calculate limits x tends to infinity. Then why we cannot use it here? Sorry I could not get your point.

[@TLMaths]

No, we don't ever write 1/infinity = 0. We write that 1/x tends to 0 as x tends to infinity. A common misconception of taking limits is thinking of it merely as substituting a value into an expression.

[@MathBuz]

TLMaths It means in situations, where it’s exact value occur, we don’t consider it.. Ok thanks 🍎

[@mohammadali-jx8ic]

Thanks mate!!

[@dansilverman6088]

What would the domain be of fg(x) if g(x) doesn't have a restricted domain? Thank you

[@TLMaths]

The domain of y=fg(x) is the intersection of the domain of y=g(x) and y=f(g(x))

[@dansilverman6088]

@@TLMaths I just finished my maths A level, thank you for always being so responsive to my comments and making excellent revision videos!

[@darenkaulsay8076]

@@TLMaths do you mean union?

[@darenkaulsay8076]

​@@TLMaths g(x) has domain of xεℝ , x≠1. f(x) has domain of xεℝ , x≠ 5/2 . fg(x) is (x-1)/(7-5x), and so has domain xεℝ , x≠7/5. So the intersection of the domains of g(x) and f(g(x)) is xεℝ But the union of the domains of g(x) and f(g(x)) is xεℝ , x≠1 , x≠7/5 ( in order to get domain of f(g(x)).

[@darenkaulsay8076]

this is the answer you gave to a previous comment, sorry for posting it here

[@veem3150]

Hi there, just a quick question please. Restating your problem, namely: f(x) = 1/x, where domain is x ∈ ℝ, x ≠ 0 and g(x) = 1/(x + 2), where domain is x ∈ ℝ, x ≠ -2 What is the domain of fg(x) which = x + 2 ? Superficially/intuitively, the domain for fg(x) would be x ∈ ℝ ; however, as you say, g(x) comes with “baggage” i.e. x ≠ -2 and hence, you state the domain of fg(x) is therefore { fg(x) : fg(x) x ∈ ℝ, x ≠ -2}. So, my daft lad question is, why isn’t the domain for fg(x) not the following i.e. { fg(x) : fg(x) x ∈ ℝ, x ≠ -2, x ≠ 0 }. In order words, why isn’t the “baggage” associated with f(x) not included (i.e. f(x) domain is x ∈ ℝ, x ≠ 0) in the domain for fg(x)? Many thanks for any steer you can provide.

[@TLMaths]

Think of the "journey" if you will of the x value you're substituting in. You start with your x value, it must be substituted into g. So your x value cannot be -2, but it can be anything else. Then, whatever value you've ended up with must be substituted into f. Whatever it is, it cannot be 0. There is no value of x you could substitute into 1/(x+2) to get zero, so there is no restriction coming from f.

[@veem3150]

Got it! Many thanks indeed.

[@veem3150]

Apologies, been thinking a little more about this. Assuming I have interpreted what you wrote correctly, then can you advise of a scenario whereby, a modified version of your posed problem, where f(x) = "something" and g(x) = "something else" would in fact necessitate the inclusion of the f(x) domain restriction within the domain of fg(x)?

[@TLMaths]

It's not that the domain of f isn't being considered, because clearly we need to check that all of the y values for y = g(x) are in the domain of f. Now, there are two ways of doing this. For example, let f(x) = 1/(2x - 5) and g(x) = 1/(x - 1). Method 1: The domain of f(x) is x ∈ ℝ, x ≠ 5/2 The domain of g(x) is x ∈ ℝ, x ≠ 1 So we cannot have g(x) = 5/2, so we find the value of x for which this occurs. 1/(x - 1) = 5/2 x - 1 = 2/5 x = 7/5 So the domain of fg(x) is x ∈ ℝ, x ≠ 1, x ≠ 7/5 Method 2 (my preferred method): The domain of g(x) is x ∈ ℝ, x ≠ 1 fg(x) = 1/(2*(1/(x - 1)) - 5) = (x - 1)/(2 - 5(x - 1)) = (x - 1)/(7 - 5x) which by itself is a function with the domain x ∈ ℝ, x ≠ 7/5 So the domain of fg(x) is x ∈ ℝ, x ≠ 1, x ≠ 7/5

[@pandaghee7524]

For gf(x) how did you get x is not equal to -1/2 I don’t understand when you set the bottom number to zero how could you get -1/2 please explain.

[@TLMaths]

We can't have the denominator being zero, so: 1 + 2x = 0 2x = -1 x = -1/2 So x cannot be equal to -1/2.