In fg(x), which function acts first?

The one written closest to xxx. So fg(x)=f(g(x))fg(x)=f(g(x))fg(x)=f(g(x)) applies ggg first, then fff. Reading the letters right to left keeps the order straight.

Functions Practice Questions

Q: Why does an absolute value equation give two answers?

Because ∣A∣=4|A|=4∣A∣=4 means AAA is 444 units from zero, so A=4A=4A=4 or A=−4A=-4A=−4. Both cases are valid, and you must solve each to find every value of xxx.

How to use these practice questions

The six questions below rise in difficulty across the whole Functions chapter, from finding an image to solving an absolute value equation. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, a method line, a clear substitution, then the final answer.

Resist the urge to peek. Only once you have committed to a full answer should you open the solution and mark yourself line by line.

When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost. Because Add Math is marked analytically, a clear method line still earns credit even when the final arithmetic slips, so always show your substitution in full.

Treat this page as a rehearsal, not a test, the point is to find weak steps now, while there is still time to fix them.

Six practice questions

Q1[3 marks]

The function $f$ is defined by $f(x)=3x-4$ . Find (a) the image of $2$ , and (b) the object whose image is $11$ .

Show worked solution

(a) The image of $2$ is the output $f(2)$ . Substitute the object $x=2$ into the rule and simplify:

f(2)=3(2)-4=6-4=2

(b) An object whose image is $11$ is a value of $x$ with $f(x)=11$ . Set the rule equal to $11$ and solve the linear equation for $x$ :

3x-4=11 \;\Rightarrow\; 3x=15 \;\Rightarrow\; x=5

Answer

The image of $2$ is $2$ , and the object is $5$ . Check part (b) by substituting back: $f(5)=3(5)-4=11$ , as required.

Q2[4 marks]

Two functions are defined by $f(x)=2x+1$ and $g(x)=x^{2}-3$ . Find (a) the composite function $fg(x)$ , and (b) the value of $gf(2)$ .

Show worked solution

(a) $fg(x)$ means $f(g(x))$ , so $g$ acts first. Replace the input of $f$ with the whole of $g(x)=x^{2}-3$ , then expand carefully:

fg(x)=f(x^{2}-3)=2(x^{2}-3)+1=2x^{2}-6+1=2x^{2}-5

(b) $gf(x)$ means $g(f(x))$ , so now $f$ acts first. Work inside-out: find $f(2)$ , then apply $g$ to that result.

f(2)=2(2)+1=5

gf(2)=g(5)=(5)^{2}-3=25-3=22

Answer

$fg(x)=2x^{2}-5$ and $gf(2)=22$ . Check part (b) the other way: $gf(x)=g(2x+1)=(2x+1)^{2}-3=4x^{2}+4x-2$ , so $gf(2)=4(4)+4(2)-2=22$ , which agrees.

Q3[4 marks]

A function is defined by $f(x)=4x+1$ . Find (a) the inverse function $f^{-1}(x)$ , and (b) the value of $f^{-1}(9)$ .

Show worked solution

(a) To find the inverse, let $y=f(x)$ , then make $x$ the subject and swap the letters at the end. Start from the rule:

y=4x+1 \;\Rightarrow\; y-1=4x \;\Rightarrow\; x=\frac{y-1}{4}

Replacing $y$ with $x$ gives the inverse function:

f^{-1}(x)=\frac{x-1}{4}

(b) Substitute $x=9$ into the inverse you found:

f^{-1}(9)=\frac{9-1}{4}=\frac{8}{4}=2

Answer

$f^{-1}(x)=\dfrac{x-1}{4}$ and $f^{-1}(9)=2$ . Check by running $f$ forward: $f(2)=4(2)+1=9$ , so the inverse sends $9$ back to $2$ , as it should.

Q4[4 marks]

The function $f$ is defined by $f(x)=3x+1$ , and the composite function $fg(x)=6x-2$ . Find the function $g(x)$ .

Show worked solution

Here $fg(x)=f(g(x))$ , so wherever $f$ has an input, that input is now the whole of $g(x)$ . Apply the rule of $f$ to $g(x)$ and set it equal to the given composite:

f(g(x))=3\,g(x)+1=6x-2

Now treat $g(x)$ as the unknown and solve. Subtract $1$ from both sides, then divide by $3$ :

3\,g(x)=6x-3 \;\Rightarrow\; g(x)=2x-1

Answer

$g(x)=2x-1$ . Check by rebuilding the composite: $fg(x)=f(2x-1)=3(2x-1)+1=6x-3+1=6x-2$ , which matches the question.

Q5[5 marks]

A function is defined by $f(x)=ax+b$ , where $a$ and $b$ are constants. Given that $f(1)=5$ and $f(3)=11$ , find (a) the values of $a$ and $b$ , and (b) the inverse function $f^{-1}(x)$ .

Show worked solution

(a) Turn each given value into an equation by substituting the object into the rule. From $f(1)=5$ and $f(3)=11$ :

a+b=5 \quad\text{and}\quad 3a+b=11

Subtract the first equation from the second to eliminate $b$ , then back-substitute:

(3a+b)-(a+b)=11-5 \;\Rightarrow\; 2a=6 \;\Rightarrow\; a=3

a+b=5 \;\Rightarrow\; 3+b=5 \;\Rightarrow\; b=2

So the function is $f(x)=3x+2$ . (b) Find the inverse by letting $y=3x+2$ and making $x$ the subject:

y=3x+2 \;\Rightarrow\; x=\frac{y-2}{3} \;\Rightarrow\; f^{-1}(x)=\frac{x-2}{3}

Answer

$a=3$ , $b=2$ , so $f(x)=3x+2$ and $f^{-1}(x)=\dfrac{x-2}{3}$ . Check the constants: $f(1)=3+2=5$ and $f(3)=9+2=11$ , both correct.

Q6[5 marks]

The function $f$ is defined by $f(x)=|\,2x-6\,|$ . Find (a) the value of $f(1)$ , and (b) all values of $x$ for which $f(x)=4$ .

Show worked solution

(a) The absolute value of a number is its distance from zero, so it is never negative. Substitute $x=1$ and take the size of the result:

f(1)=|\,2(1)-6\,|=|-4|=4

(b) The equation $|\,2x-6\,|=4$ means the inside expression is $4$ units from zero, so it can equal either $+4$ or $-4$ . Split into two cases and solve each:

2x-6=4 \;\Rightarrow\; 2x=10 \;\Rightarrow\; x=5

2x-6=-4 \;\Rightarrow\; 2x=2 \;\Rightarrow\; x=1

Answer

$f(1)=4$ , and $f(x)=4$ when $x=1$ or $x=5$ . Check both: $|\,2(1)-6\,|=|-4|=4$ and $|\,2(5)-6\,|=|4|=4$ .

Splitting into a positive and a negative case is what makes sure you find both answers, not just one.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct method being started, the right substitution, and a clean final statement.

Method mark: did you write the correct rule or move, for example $fg(x)=f(g(x))$ , or setting the rule equal to the given image?
Substitution mark: is the correct value put into the correct place, with brackets around any negative object?
Answer mark: is the final value stated clearly, and does it survive a check by substituting back?
For an absolute value equation, you only earn full marks if you show both the positive and the negative case.
If your final number is wrong but the method line is right, give yourself the method mark, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a composite read in the wrong order, a missing bracket, or only one case shown for an absolute value equation, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How long should each of these questions take me?

Aim for roughly five to eight minutes each, rising with the mark value. If a question takes far longer, note it and bring it to a lesson, the time it steals in the exam is often the real problem, not the topic itself.

fg(x)

, which function acts first?

The one written closest to $x$ . So $fg(x)=f(g(x))$ applies $g$ first, then $f$ .

Reading the letters right to left keeps the order straight.

Why does an absolute value equation give two answers?

Because $|A|=4$ means $A$ is $4$ units from zero, so $A=4$ or $A=-4$ . Both cases are valid, and you must solve each to find every value of $x$ .

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct method line and a correct substitution still earn marks even if the arithmetic slips at the end.

That is why you should always show full working.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Functions, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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