smc-642-40-Other functions

Functions, MET1 2023 VCE SM-Bank 2

Consider the functions \(f\) and \(g\), where

\begin{aligned}
& f: R \rightarrow R, f(x)=x^2-9 \\
& g:[0, \infty) \rightarrow R, g(x)=\sqrt{x}
\end{aligned}

State the range of \(f\). (1 mark)

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Determine the rule for the equation and state the domain of the function \(f \circ g\). (2 marks)

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Let \(h\) be the function \(h: D \rightarrow R, h(x)=x^2-9\).
Determine the maximal domain, \(D\), such that \(g \circ h\) exists. (1 marks)

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a. \([-9, \infty)\)

b. \(f\circ g(x)=x-9, \text{Domain}\ [0, \infty)\)

c. \((-\infty, -3)\cap (3, \infty)\)

Show Worked Solution

a. \(\text{Range}\ \rightarrow\ [-9, \infty)\)

b.	\(f\circ g(x)\)	\(=(g(x))^2-9\)
		\(=(\sqrt{x})^2-9\)
		\(=x-9\)

\(g(x)=\sqrt{x} \ \rightarrow x\ \text{must be }\geq 0\)

\(\therefore\ \text{Domain}\ f\circ g(x) \text{ is }[0, \infty)\)

c.	\(g\circ h(x)\)	\(=\sqrt{h(x)}\)
		\(=\sqrt{x^2-9}\)

\(\text{For }g\circ h(x)\ \text{to exist}\ h(x)\geq 0\)

\(x\text{-intercepts for }h(x)\ \text{are } x=-3, 3\)

\(\text{and }h(x)\ \text{is positive for } x\leq -3\ \text{and }x\geq 3\)

\(\therefore\ \text{Maximal domain} = (-\infty, -3)\cap (3, \infty)\)

Algebra, MET2 2023 VCAA 16 MC

Let \(f(x)=e^{x-1}\).

Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is

\(g(x)=e^{x-1}\)
\(g(x)=e^{(x-2)(x-1)}\)
\(g(x)=e^{(x+2)(x-1)}\)
\(g(x)=e^{x(x-2)}\)
\(g(x)=e^{x(x-3)}\)

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\(B\)

Show Worked Solution

\(f(x)\times g(x)\)	\(=e^{(x-1)^2}\)
\(e^{x-1}\times g(x)\)	\(=e^{(x-1)^2}\)
\( g(x)\)	\(=\dfrac{e^{(x-1)^2}}{e^{(x-1)}}\)
	\(=e^{(x-1)^2}\times e^{(x-1)^-1}\)
	\(=e^{x^2-3x+2}\)
	\(=e^{(x-2)(x-1)}\)

\(\Rightarrow B\)

Graphs, MET2 2023 VCAA 3 MC

Two function, \(p\) and \(q\), are continuous over their domains, which are \([-2, 3)\) and \((-1, 5]\), respectively.

The domain of the sum function \(p+q\) is

\([-2, 5]\)
\([-2, -1)\cup (3, 5]\)
\([-2, -1)\cup (-1, 3)\cup(3, 5]\)
\([-1, 3]\)
\((-1, 3)\)

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\(E\)

Show Worked Solution

\(\text{Domain of sum function = intersection of two domains.}\)

\([-2, 3)\cap(-1, 5]=(-1, 3)\)

\(\Rightarrow E\)

♦ Mean mark 47%.

Functions, MET2 2021 VCAA 10 MC

Consider the functions `f(x) = sqrt{x+2}` and `g(x) = sqrt{1-2x}`, defined over their maximal domains.

The maximal domain of the function `h = f + g` is.

`(–2, 1/2)`
`[–2,∞)`
`(–∞, –2) ∪ (1/2, ∞)`
`[–2, 1/2]`
`[–2, 1]`

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`D`

Show Worked Solution

`f(x) \ = \ sqrt{x + 2} \ => \ text{domain} \ x ≥ -2`

`g(x) \ = \ sqrt{1-2x} \ => \ text{domain} \ x ≤ 1/2`

`text{Intersection of domains = domain} \ h(x)`

`:. \ h(x) ∈ [-2, 1/2]`

`=> D`

Algebra, MET1 2019 VCAA 8

The function `f: R -> R, \ f(x)` is a polynomial function of degree 4. Part of the graph of `f` is shown below.

The graph of `f` touches the `x`-axis at the origin.

Find the rule of `f`. (1 mark)

Let `g` be a function with the same rule as `f`.

Let `h: D -> R, \ h(x) = log_e (g(x)) - log_e (x^3 + x^2)`, where `D` is the maximal domain of `h`.

State `D`. (1 mark)
State the range of `h`. (2 marks)

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`f(x) = -4x^2(x^2 – 1)`
`x in (-1, 1) text(\{0})`
`h(x) in (-oo, 3 log_e 2) text(\ {)2 log_e 2 text(})`

Show Worked Solution

a.	`y`	`= ax^2 (x – 1)(x + 1)`
		`= ax^2 (x^2 – 1)\ …\ (1)`

`text(Substitute)\ (1/ sqrt 2, 1)\ text{into (1):}`

`1 = a ⋅ (1/sqrt 2)^2 ((1/sqrt 2)^2 – 1)`

`1 = a(1/2)(-1/2)`

`a = -4`

`:. f(x) = -4x^2(x^2 – 1)`

b.	`g(x) > 0`	`=> -4x^2 (x^2 – 1) > 0`
		`=> x in (-1, 1)\ text(\{0})`

`text(and)`

`x^3 + x^2 > 0 => text(true for)\ \ x in (-1 , 1)\ text(\{0})`

`:. D:\ x in (-1, 1)\ text(\{0})`

c.	`h(x)`	`= log_e ((-4x^2(x^2 – 1))/(x^3 + x^2))`
		`= log_e ((-4x^2(x + 1)(x – 1))/(x^2(x + 1)))`
		`= log_e (4(1 – x))\ \ text(where)\ \ x in (-1, 1)\ text(\{0})`

`text(As)\ \ x -> -1,\ \ h(x) -> log_e 8 = 3 log_e 2`

`text(As)\ \ x -> 1,\ \ h(x) -> -oo`

`text(As)\ \ x -> 0,\ \ h(x) -> log_e 4 = 2 log_e 2`

`text{(}h(x)\ text(undefined when)\ \ x = 0 text{)}`

`:.\ text(Range)\ \ h(x) in (-oo, 3 log_e 2)\ text(\{) 2 log_e 2 text(})`

Algebra, MET2 2018 VCAA 10 MC

The function `f` has the property `f (x + f (x)) = f (2x)` for all non-zero real numbers `x`.

Which one of the following is a possible rule for the function?

`f(x) = 1 - x`
`f(x) = x - 1`
`f(x) = x`
`f(x) = x/2`
`f(x) = (1 - x)/2`

Show Answers Only

`C`

Show Worked Solution

`text(By trial and error,)`

`text(Consider option C:)`

`x + f(x)`	`= x + x = 2x`
`f(2x)`	`=2x`
`:. f(x + f(x))`	`= f(2x)`

`=> C`

Algebra, MET2 2017 VCAA 13 MC

Let `h:(−1,1) -> R`, `h(x) = 1/(x - 1)`.

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

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

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`E`

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`

`=> E`

Algebra, MET2 2008 VCAA 12 MC

Let `f: R -> R,\ f(x) = e^x + e^(–x).`

For all `u in R,\ f(2u)` is equal to

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

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`C`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 44%.

`text(Define)\ \ f(x) = e^x + e^-x`

`text(Enter each functional equation)`

`[text(i.e.)\ \ f(2u) = (f(u))^2 – 2]`

`text(until CAS output is “true”)`

`=> C`

`text(Solution 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`

Algebra, MET2 2009 VCAA 5 MC

Let `f: R -> R,\ 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)`
`f (x + y) + f (x - y) = 2 (f (x) + f(y))`

Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`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`

`text(Solution 2)`

`text(Define)\ \ f(x) = x^2`

`text(Enter each functional equation on CAS)`

`text(until output does NOT read “true”.)`

`=> D`

Algebra, MET2 2016 VCAA 11 MC

The function `f` has the property `f(x) - f(y) = (y - x)\ f(xy)` for all non-zero real numbers `x` and `y`.

Which one of the following is a possible rule for the function?

`f(x) = x^2`
`f(x) = x^2 + x^4`
`f(x) = x log_e (x)`
`f(x) = 1/x`
`f(x) = 1/x^2`

Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 47%.

`text(Consider option)\ D:`

`text(LHS)\ = 1/x – 1/y`

`text(RHS)\ =(y-x) xx 1/(xy) = 1/x – 1/y =\ text(LHS)`

`=>D`

`text{Solution 2 (using technology)}`

`text(Define each specific function on CAS)`

`[text(i.e.)\ \ f(x) = 1/x]`

`text(Enter functional equation)\ \ f(x) – f(y) = (y – x)\ f(xy)`

`text(unitl CAS output is “TRUE”)`

`=> D`

Algebra, MET2 2013 VCAA 13 MC

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

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

Show Answers Only

`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)`

`text(CAS can be used to quickly prove other answers)`

`text(do not satisfy equation.)`

`=> C`

Algebra, MET2 2013 VCAA 5 MC

If `f: text{(−∞, 1)} -> R,\ \ f(x) = 2 log_e (1 - x)\ \ text(and)\ \ g: text{[−1, ∞)} -> R, g(x) = 3 sqrt (x + 1),` then the maximal domain of the function `f + g` is

`text{[−1, 1)}`
`(1, oo)`
`text{(−1, 1]}`
`text{(−∞, −1]}`
`R`

Show Answers Only

`A`

Show Worked Solution

`text(Consider)\ \ f(x) = 2 log_e (1 – x):`

`(1-x)`	`>0`
`:. x`	`<1`

`text(Consider)\ \ g(x) = 3 sqrt (x + 1):`

`(x+1)`	`>=0`
`:. x`	`>= -1`

`:.\ text(The maximal domain of)\ \ f + g\ \ text{is [−1, 1)}.`

`=> A`