Composite Functions

Functions, 2ADV EQ-Bank 11

Given the functions \(f(x)=x-2\) and \(g(x)=x^2\), determine the equation of \(f(g(x))\) and sketch its graph showing all intercepts. (2 marks)

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Evaluate \(g(f(-1))\). (1 mark)

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b. \(g(f(-1))=9\)

Show Worked Solution

a. \(f(x)=x-2, \ g(x)=x^2\)

\(f(g(x))=x^2-2\)

\(\text{Intercepts:} \ \ x^2-2=0 \ \Rightarrow \ x= \pm \sqrt{2}\)

b. \(f(-1)=-1-2=-3\)

\(g(f(-1))=g(-3)=9\)

Functions, 2ADV F1 2025 HSC 18

Find the range of \(g(f(x))\), given \(f(x)=\dfrac{3}{x-1}\) and \(g(x)=x+5\). (2 marks)

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\(\text{Range} \ g(f(x)): \text{All} \ y, \ y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Show Worked Solution

\(f(x)=\dfrac{3}{x-1}, \ \ g(x)=x+5\)

\(g(f(x))=\dfrac{3}{x-1}+5 \Rightarrow \ \text{vertical translation +5 of} \ f(x)\)

\(\text{Range} \ f(x):\ \text{All} \ y, \ y \neq 0\)

\(\text{Range} \ g(f(x)): \text{All} \ y, \ y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Functions, 2ADV F1 EQ-Bank 19

Given that \(f(x)=x^2+1\) and \(g(x)=x+2\), determine \(f(g(x))\) and its range. (2 marks)

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\(f(g(x))=(x+2)^2+1\)

\(\ \text{Range} \ f(g(x)): \ y \geqslant 1\)

Show Worked Solution

\(f(x)=x^2+1, \quad g(x)=x+2\)

\(f(g(x))=(x+2)^2+1\)

\(\Rightarrow f(g(x)) \ \text{is a concave up parabola, vertex}\ (-2,1)\)

\(\therefore \ \text{Range} \ f(g(x)): \ y \geqslant 1\)

Functions, 2ADV F1 EQ-Bank 18

If \(f(x)=x^2-3\) and \(g(x)=\sqrt{x-2}\),

Find \(f(g(x))\). (1 mark)
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What is the domain of \(f(g(x))\) ? (1 mark)
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a. \(f(g(x))=x-5\)

b. \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\)

\(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)

Show Worked Solution

a.	\(f(g(x))\)	\(=(\sqrt{x-2})^2-3\)
		\(=x-2-3\)
		\(=x-5\)

b. \(\text{Restriction for}\ \ g(x)\ \Rightarrow \ x-2\geqslant 0 \ \Rightarrow \ x \geqslant 2\)

\(\text{Domain}\ \ f(g(x)):\ x \geqslant 2\)

Functions, 2ADV F1 2022 HSC 10 MC

The graphs of `y=f(x)` and `y=g(x)` are shown.

Which graph best represents `y=g(f(x))`

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`B`

Show Worked Solution

`text{By Elimination:}`

`y=f(x)\ \ text{is an even function}`

`y=f(x)=f(-x)\ \ =>\ \ g(f(x))=g(f(-x))`

`g(f(x))\ \ text{is also an even function}`

`text{→ Eliminate A, C and D}`

`=>B`

♦ Mean mark 46%.

Functions, 2ADV F1 EQ-Bank 4 MC

Let `f(x)` and `g(x)` be functions such that `f(–1)=4, \ f(2)=5, \ g(–1)=2, \ g(2)=7` and `g(4)=6`.

The value of `g(f(–1))` is

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`D`

Show Worked Solution

`f(–1)=4`

`g(f(–1)) = g(4) = 6`

`=>D`

Functions, 2ADV F1 EQ-Bank 20

Given the function `f(x) = sqrt(3-x)` and `g(x) = x^2-2`, sketch `y = g(f(x))` over its natural domain. (2 marks)

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Show Worked Solution

`g(x) = x^2-2,\ \ f(x) = sqrt(3-x)`

`g(f(x))`	`= (sqrt(3-x))^2-2`
	`= 3-x-2`
	`= 1-x`

`text(S)text(ince)\ \ f(x) = sqrt(3-x),`

`=> text(Domain:)\ x <= 3`

Functions, 2ADV F1 2019 MET1-N 2

Let `f(x) = -x^2 + x + 4` and `g(x) = x^2-2`.

Find `g(f(3))`. (2 marks)

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Express `f(g(x))` in the form `ax^4 + bx^2 + c`, where `a`, `b` and `c` are non-zero integers. (2 marks)

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a. `2`

b. `-x^4 + 5x^2-2`

Show Worked Solution

a. `f(3)= -3^2 + 3 + 4= -2`

`g(f(3))= g(-2)= (-2)^2-2= 2`

b. `f(g(x))`	`= -(x^2-2)^2 + (x^2-2) + 4`
	`= -(x^4-4x^2 + 4) + x^2 + 2`
	`= -x^4 + 5x^2-2`

Functions, 2ADV F1 EQ-Bank 21

Find the domain and range of `f(g(x))` given

`f(x) = 2x^2 - 8x` and `g(x) = x + 2`. (2 marks)

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`text(Domain: all)\ x`

`text(Range:)\ −8<=y< ∞`

Show Worked Solution

`f(g(x))`	`= 2(x + 2)^2 – 8(x + 2)`
	`= 2(x^2 + 4x + 4) – 8x – 16`
	`= 2x^2 + 8x + 8 – 8x – 16`
	`= 2(x^2 – 4)`

`:. text(Domain: all)\ x`

`:. text(Range:)\ −8<=y< ∞`

Functions, 2ADV F1 EQ-Bank 17

Given `f(x) = sqrtx` and `g(x) = 25-x^2`

Find `g(f(x))`. (1 mark)

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Find the domain and range of `f(g(x))`. (2 marks)

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a. `25-x`

b. `text(Domain:)\ −5<= x <= 5`

`text(Range:)\ 0<=y<= 5`

Show Worked Solution

a. `g(f(x))= 25-(f(x))^2= 25-(sqrtx)^2= 25-x`

b. `f(g(x))= sqrt(g(x))= sqrt(25-x^2)`

`:.\ text(Domain:)\ −5<= x <= 5`

`:.\ text(Range:)\ 0<=y<= 5`

Functions, 2ADV F1 EQ-Bank 9 MC

If the equation `f(2x)-2f(x) = 0` is true for all real values of `x`, then `f(x)` could equal

`x^2/2`
`sqrt (2x)`
`2x`
`x-2`

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`C`

Show Worked Solution

`text(We need)\ \ f(2x)=2\ f(x),`

`text(Consider)\ C:`

`f(x)`	`=2x,`
`f(2x)`	`= 2(2x)= 2\ f(x)`

`=> C`

Functions, 2ADV F1 EQ-Bank 14 MC

Let `g(x) = log_2(x),\ \ x > 0`

Which one of the following equations is true for all positive real values of `x?`

`2g (8x) = g (x^2) + 8`
`2g (8x) = g (x^2) + 6`
`2g (8x) = (g (x) + 8)^2`
`2g (8x) = g (2x) + 6`

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`B`

Show Worked Solution

`text(Consider Option)\ B:`

♦♦ Mean mark 35%.

`text(LHS)`	`= 2g(8x)`
	`= 2log_2(8x)`
	`= 2log_2(8) + 2log_2(x)`
	`=2log_2 (2^3)+ 2log_2(x)`
	`= 6 + log_2(x^2)`
	`= g(x^2) + 6`

`=> B`

Functions, 2ADV F1 EQ-Bank 6 MC

Which one of the following functions satisfies the functional equation `f (f(x)) = x`?

A. `f(x) = 2 - x`

B. `f(x) = x^2`

C. `f(x) = 2 sqrt x`

D. `f(x) = x - 2`

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`A`

Show Worked Solution

`text(By trial and error,)`

`text(Consider:)\ \ f(x)=2-x`

`f(f(x))`	`=2-(2-x)`
	`=x`

`=> A`

Functions, 2ADV F1 EQ-Bank 8 MC

If `f(x - 1) = x^2 - 2x + 3`, then `f(x)` is equal to

A. `x^2 - 2`

B. `x^2 + 2`

C. `x^2 - 2x + 4`

D. `x^2 - 4x + 6`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ g(x) = f(x – 1)`

`g(x+1) = f(x)`

`g(x + 1)`	`= (x + 1)^2 – 2(x + 1) + 3`
	`=x^2+2x+1-2x-2+3`
	`= x^2 + 2`

`=> B`

Functions, 2ADV F1 EQ-Bank 30

Given `f(x) = sqrt (x^2 - 9)` and `g(x) = x + 5`

Find integers `c` and `d` such that `f(g(x)) = sqrt {(x + c) (x + d)}` (2 marks)

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State the domain for which `f(g(x))` is defined. (2 marks)

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`c = 2, d = 8 or c = 8, d = 2`
`x in (– oo, – 8] uu [– 2, oo)`

Show Worked Solution

a.	`f(g(x))`	`= sqrt {(x + 5)^2 – 9}`
		`= sqrt (x^2 + 10x + 16)`
		`= sqrt {(x + 2) (x + 8)}`

`:. c = 2, d = 8 or c = 8, d = 2`

b. `text(Find)\ x\ text(such that:)`

♦♦♦ Mean mark 13%.
MARKER’S COMMENT: “Very poorly answered” with a common response of `-3 <=x<=3` that ignored the information from part (a).

`(x+2)(x+8) >= 0`

`(x + 2) (x + 8) >= 0\ \ text(when)`

`x <= -8 or x >= -2`

`:.\ text(Domain:)\ \ x<=-8\ \ and\ \ x>=-2`

Functions, 2ADV F1 EQ-Bank 11

Let `f(x) = x^2 + 1 and g(x) = 2x + 1.` Write down the rule of `f(g(x)).` (1 mark)

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`(2x + 1)^2 + 1`

Show Worked Solution

`f (g(x))=f(2x+1)= (2x + 1)^2 + 1`

Functions, 2ADV F1 EQ-Bank 3 MC

If `f(x) = 1/2e^(3x) and g(x) = log_e(2x) + 3` then `g (f(x))` is equal to

`3(x + 1)`
`e^(3x) + 3`
`e^(8x + 9)`
`log_e (3x) + 3`

Show Answers Only

`A`

Show Worked Solution

`g(f(x))`	`= log_e(2 xx 1/2e^(3x)) + 3`
	`=log_e e^(3x) + 3`
	`=3x + 3`
	`= 3 (x + 1)`

`=> A`

Functions, 2ADV F1 EQ-Bank 7 MC

Let `f (x) = x^2`

Which one of the following is not true?

`f(xy) = f (x) f (y)`
`f(x)-f(-x) = 0`
`f (2x) = 4 f (x)`
`f (x-y) = f(x)-f(y)`

Show Answers Only

`D`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option)\ D:`

`f(x-y)`	`=(x-y)^2=x^2 -2xy+y^2`
`f(x)-f(y)`	`= x^2-y^2`

`:.f(x-y) !=f(x)-f(y)`

`=>D`

Functions, 2ADV F1 EQ-Bank 26

Let `f(x) = log_e(x)` for `x>0,` and `g (x) = x^2 + 1` for all `x`.

Find `h(x)`, where `h(x) = f (g(x))`. (1 mark)

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State the domain and range of `h(x)`. (2 marks)

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Show that `h(x) + h(−x) = f ((g(x))^2 )`. (2 marks)
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a. `log_e(x^2 + 1)`

b. text(Domain)\ (h):\ text(all)\ x`

`text(Range)\ h(x):\ \ h>=0`

c. `text(See Worked Solutions)`

Show Worked Solution

a.	`h(x)`	`= f(x^2 + 1)`
		`= log_e(x^2 + 1)`

b. `text(Domain)\ (h) =\ text(Domain)\ (g):\ text(all)\ x`

♦♦ Mean mark part (a)(ii) 30%.

`=> x^2 + 1 >= 1`

`=> log_e(x^2 + 1) >= 0`

`:.\ text(Range)\ h(x):\ \ h>=0`

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.

c.	`text(LHS)`	`= h(x) + h(−x)`
		`= log_e(x^2 _ 1) + log_e((−x)^2 + 1)`
		`= log_e(x^2 + 1) + log_e(x^2 + 1)`
		`= 2log_e(x^2 + 1)`

`text(RHS)= f((x^2 + 1)^2)2log_e(x^2 + 1)`

`:. h(x) + h(-x) = f((g(x))^2)\ \ text(… as required)`

Functions, 2ADV F1 EQ-Bank 15 MC

Let `f(x) = e^x + e^(-x).`

`f(2u)` is equal to

`f(u) + f(-u)`
`2 f(u)`
`(f(u))^2-2`
`(f(u))^2 + 2`

Show Answers Only

`C`

Show Worked Solution

`text(By trial and error,)`

♦ Mean mark 44%.

`text(Consider)\ \ (f(u))^2-2:`

`f(2u)`	`=e^(2u) + e^(-2u)`
`(f(u))^2`	`=(e^u + e^(-u))^2`
	`=e^(2u) + 2 + e^(-2u)`

`:. f(2u) = (f(u))^2-2`

`=>C`

Functions, 2ADV F1 EQ-Bank 1 MC

Let `g(x) = x^2 + 2x-3` and `f(x) = e^(2x + 3).`

Then `f(g(x))` is given by

`e^(4x + 6) + 2 e^(2x + 3)-3`
`2x^2 + 4x-6`
`e^(2x^2 + 4x-3)`
`e^(2x^2 + 4x-6)`

Show Answers Only

`C`

Show Worked Solution

`text(By trial and error,)`

`text(Consider:)\ \ f(x) = e^(2x^2 + 4x-3)`

`f(g(x))`	`=e^(2 (x^2 + 2x-3)+3)`
	`= e^(2x^2 + 4x-3)`

`=> C`

Functions, 2ADV F1 EQ-Bank 13 MC

The function `f(x)` satisfies the functional equation `f (f (x)) = x` for `{x:\ text(all)\ x,\ x!=1}`.

The rule for the function is

`f(x) = x + 1`
`f(x) = x-1`
`f(x) = (x-1)/(x + 1)`
`f(x) = (x + 1)/(x-1)`

Show Answers Only

`D`

Show Worked Solution

`text(By trial and error:)`

♦ Mean mark 47%.

`text(Consider)\ \ f(x) = (x + 1)/(x-1)`

`f(f(x))`	`=((x + 1)/(x-1) +1)/((x + 1)/(x-1) -1)`
	`=(x+1+x-1)/(x+1-x+1)`
	`=x`

`=>D`

Functions, 2ADV F1 EQ-Bank 25

Let `f(x) = sqrt(x + 1)` for `x>=0`

State the range of `f(x)`. (1 mark)

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Let `g(x)=x^2+4x+3`, where `x<=c` and `c<=0`.
Find the largest possible value of `c` such that the range of `g(x)` is a subset of the domain of `f(x)`. (2 marks)

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a. `y>= 1`

b. `-3`

Show Worked Solution

a. `text(Sketch of)\ \ f(x):`

`:.\ text(Range:)\ \ y>= 1`

b. `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`

`text(Domain of)\ \ f(x):\ \ x>=0`

`text(Find domain of)\ g(x)\ text(such that range)\ g(x):\ y>=0`

`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`

`x <= –3\ \ text(and)\ \ x>=-1`

`:. c = -3`

Functions, 2ADV F1 EQ-Bank 2 MC

Let `f(x)` and `g(x)` be functions such that `f (2) = 5`, `f (3) = 4`, `g(2) = 5`, `g(3) = 2` and `g(4) = 1`.

The value of `f (g(3))` is

Show Answers Only

`D`

Show Worked Solution

`f(g(3))=f(2)=5`

`=> D`

Functions, 2ADV F1 EQ-Bank 12 MC

Let `h(x) = 1/(x - 1)` for `-1<h<1`.

Which one of the following statements about `h` is not true?

`h(x)h(–x) = –h(x^2)`
`h(x) - h(0) = xh(x)`
`h(x) - h(–x) = 2xh(x^2)`
`(h(x))^2 = h(x^2)`

Show Answers Only

`D`

Show Worked Solution

`text(By trial and error, consider option)\ E:`

♦ Mean mark 46%.

`h(x) = 1/(x – 1)`

`(h(x))^2 = 1/(x – 1)^2=1/(x^2-2x+1)`

`h(x^2)=1/(x^2-1) != (h(x))^2`

`=> D`