For gf to exist, the range of f must be a subset of the domain of g.

Rf ⊂ Dg

(The range of f is basically all the possible values of f(x), the dependent variable

We are using the values of f(x) and substituting them into g(X), where X = f(x). X here is the independent variable.

If the values of f(x) are not within the domain of g, then they are outside the definition of this function.

We can't even plot them, let alone find the dependent variable gf(x) )



The range of f , Rf = (-1,1]

The domain of g, Dg = (-2, 2]

Since Rf ⊂ Dg , gf exists.


(Represent this on number line and we can clearly see that the Dg encompasses Rf entirely)

Hence, we use the range of f, (-1, 1] as the independent variables /x-values . We substitute them into the function g.


g : x↦(2x + 1)²/³ , x ∈ R, -2 < x ≤ 2


When x = -1,

g(x) = (-2 + 1)²/³
= (-1)²/³
= (³√-1)²
= (-1)²
= 1

(Real-valued root is taken here. There are also complex roots of ³√-1 which we do not consider)

When x = 1,
g(x) = (2 + 1)²/³
= 3²/³ (Or ³√9)


Now you would think that the range of gf is (1, 3²/³]
(curved brackets for 1 since the domain doesn't actually include x = -1)


But realise that (2x + 1)²/³ = (³√(2x + 1))² .
And the square of any real value is always bigger or equal to 0.

i.e (³√(2x + 1))² ≥ 0 for all x ∈ R


So the smallest possible value of (2x + 1)²/³
is actually 0. And this occurs at x = -½.

(2(-½) + 1)²/³
= (-1 + 1)²/³
= 0²/³
= 0

Since x = -½ is within the domain used (-1,1] this means the range of gf would be [0,3²/³]

For parts like (IV) should use GC. No point risking the marks thinking what’s the lowest/maximum values for the given domain since substituting the min/max X-values doesn’t work sometimes. (Sure for a straight line you can but for functions with min/max/reflection about X-axis similar to this case, will be conceptually wrong)

Plot (2x+1)^2/3 into your GC and then look at the values from (-1,1]. From GC is very clear that the curve goes down from X=-1 to X=-0.5 before going up again. The range of said function is therefore 0 at X=-0.5 and 2.08 at X=1.

Remember to diligently use your GC (especially) when it comes to Graphs and Functions unless otherwise stated.

Thank you so much for all the replies! I get it now :)

From part iii) already should use. It is more difficult to solve algebraically and manually.

However, if you really want to understand why the minimum is 0, it is through the algebraic demonstration.

The graph plotted is afterall based on the algebraic meaning that we derive.

If we understand that the minimum is 0 just because 'the graph shows it', we are missing out on the essence.

Besides , GC will give you 2.08 at x = 2, which is less accurate than 3²/³ when calculated with workings.

So in exam, do a combination of both. Use the GC to get the graph for visualisation, but for the range try to get the exact answer with manual calculation.

Also , if you are asked to justify why the minimum is 0, the algebraic working I provided above will come in handy.

I can only say GC is more of a exam tool /technique. The fundamental understanding should still be learnt

Additional algebraic explanation for why the maximum is 3²/³ :

(2x + 1)²/³ = ³√(2x + 1)²

(2x + 1)² > 0 for x > -½ and x < -½, since |2x + 1| will be > 0 for those ranges.

Squaring gives a positive value so the bigger the absolute value/modulus, the bigger the square.


Now (2x + 1)² > 0 implies ³√(2x + 1)² > 0
(The cube root of a positive real value is always positive)

So the function will strictly increase when x decreases from -½ onwards (not just to -1) or increases from -½ onwards(not just to 1) since |2x + 1| can only increase from then.

The function is symmetric about x = -½.


Since x = 1 is further away from x = -½ than x = -1 is, this means |2x + 1| is bigger and so will (2x + 1)²/³ be.

i.e |2(1) + 1| = |3| > |2(-1) + 1| = |1|


From this we know that x = 1 will be the maximum for the domain given.


Students should know this just in case they're asked to find without use of a GC. From how I see it, Hwa Chong setters are trying to test their students on the how deep their understanding of a function's range is based on its definition

Thank you so much for all the explanation :D i always struggle with finding the range of composite functions without a gc haha

Welcome.

The whole idea of my explanation is to help students to see things from the fundamentals, instead of taking the easy way out and using online software like Desmos or our GC. Sure, we can use them, but to check or verify if we got our concepts right. Relying too much on software limits our understanding and learning.

It's just like PSLE students nowadays, overelying on their calculators , and many struggle with their Math Paper 1 where calculators are not allowed. The problem lies in not having developed their mental and written calculation skills properly.

In exam, time constraints apply and it's not easy to come up with a robust explanation if GC is not allowed. The better you understands the fundamentals, the faster you'll come up with it.

Also we notice from this example that we don't actually need to find gf(x) to solve iv)

just curious, if i want to get my fundementals right, is continuous practice the key? Because i often hear that for math it's important to keep practising, but i feel that im not improving even tho im doing more qns :/

I would suggest to read up more on the topics. There are many good free math resources online covering topics in H2 Math, eg those from universities. There are also some local H2 Math websites that do have post addressing past year questions with analysis and solutions. Also, consult your JC tutors often on any doubts or uncertainty that you have, no matter how simplistic or trivial you think the questions may be.

Continuous practice is only useful once we have understood what the topics are all about. Practice makes permanent, not perfect. Continuous practice is merely to make what you have been asked to learn into second nature.

But the trend of A level exams nowadays is to really grill and test students conceptually. Questions are getting creative and abstract. Practising 10 years series and school papers repeatedly doesn't guarantee an A nowadays.

One example would be asking a student to complete the square but with all constants being alphabets. Another would be to give you a question on graphing techniques but in the form y = (x - r)/(x + p) and asking you questions about r and p, or the derivative. Without numbers, things get very abstract and this is where we go back to basics.


The problem with JC education is that too many topics are covered in too little time. Teachers tend to gloss over detailed explanation and just push formulas.

Eg. Things like dy/dx, z = re^iθ, function range and domain, cross product, a.b = |a||b| cos θ and maclaurins series are easily performed. But how to understand them intuitively? These they don't cover in depth and sometimes they make no sense.


So a lot of work depends on our own effort to try to get a better grasp of what we're being taught.

So in short, continue to practise, but as you do so, keep reading up on these topics. Another good thing is to ask your teachers to explain solutions in detail. Hopefully they have the time.

Oh okay! :) thanks so much for helping all this while im really really grateful