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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}

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

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

  3. Let \(h\) be the function \(h: D \rightarrow R, h(x)=x^2-9\).
  4. 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.     \(g\circ h(x)\) \(=\sqrt{h(x)}\)
    \(=\sqrt{x^2-9}\)

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

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

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

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

Graphs, MET2 2022 VCAA 4 MC

Which one of the following functions is not continuous over the interval `x \in[0,5]`?

  1. `f(x)=\frac{1}{(x+3)^2}`
  2. `f(x)=\sqrt{x+3}`
  3. `f(x)=x^{\frac{1}{3}}`
  4. `f(x)=\tan \left(\frac{x}{3}\right)`
  5. `f(x)=\sin ^2\left(\frac{x}{3}\right)`
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`D`

Show Worked Solution

From graph `f(x)=\tan \left(\frac{x}{3}\right)` is not continuous for `x \in[0,5]`

Alternatively `f(x)=\tan \left(\frac{x}{3}\right)` is discontinuous when `x/3 = pi/2`

i.e. when `x = (3pi)/2 ~~ 4.71… <5`

 

`=>D`

Graphs, MET2 2020 VCAA 18 MC

Let `a \in(0, \infty)` and `b \in R`.

Consider the function  `h:[-a, 0) \cup(0, a] \rightarrow R, h(x)=\frac{a}{x}+b`.

The range of  `h`  is

  1. `[b-1,b+1]`
  2. `(b-1,b+1)`
  3. `(-oo,b-1)uu(b+1,oo)`
  4. `(-oo,b-1]uu[b+1,oo)`
  5. `[b-1,oo)`
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`D`

Show Worked Solution

`h(x)=(a)/(x)+b`

♦ Mean mark 43%.
MARKER’S COMMENT: Use  `y=2/x-3`  to illustrate an example of endpoints etc..

`text(Graph will be in the shape of a hyperbola)`

`text(Endpoint coordinates:)`

`(-a,-1+b) and (a, 1+b)`

`:.\ “Range”=(-oo,-1+b]uu[b+1,oo)`

`=>D`

Algebra, MET2 2018 VCAA 3 MC

Consider the function  `f: [a, b) -> R,\ f(x) = 1/x`, where `a` and `b` are positive real numbers.

The range of  `f` is

  1. `[1/a, 1/b)`
  2. `(1/a, 1/b]`
  3. `[1/b, 1/a)`
  4. `(1/b, 1/a]`
  5. `[a, b)`
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`D`

Show Worked Solution

`f: [a, b) -> R,\ f(x) = 1/x`

♦ Mean mark 48%.

`f(a) = 1/a,\ f(b) = 1/b`
 

`text(S) text(ince)\ a < b,`

`=>\ f(a) > f(b)`

`:.\ text(Range)\ \ f: (1/b, 1/a]`

`=>   D`

Graphs, MET2 2018 VCAA 2 MC

The maximal domain of the function  `f`  is  `R text(\{1})`.

A possible rule for  `f` is

  1. `f(x) = (x^2 - 5)/(x - 1)`
  2. `f(x) = (x + 4)/(x - 5)`
  3. `f(x) = (x^2 + x + 4)/(x^2 + 1)`
  4. `f(x) = (5 - x^2)/(1 + x)`
  5. `f(x) = sqrt (x - 1)`
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`A`

Show Worked Solution

`text(Maximal domain of)\ \ f -> R text(\{1})`

`->\ text(all real)\ x,\ x != 1`

`:. f(x) = (x^2 – 5)/(x – 1)`

`=>   A`

Graphs, MET2 2008 VCAA 8 MC

The graph of the function  `f: D -> R,\ f(x) = (x - 3)/(2 - x),` where `D` is the maximal domain has asymptotes

  1. `x = 3,\ \ \ \ \ \ \ \ \ \ y = 2`
  2. `x = -2,\ \ \ \ \ y = 1`
  3. `x = 1,\ \ \ \ \ \ \ \ \ \ y = -1`
  4. `x = 2,\ \ \ \ \ \ \ \ \ \ y = -1`
  5. `x = 2,\ \ \ \ \ \ \ \ \ \ y = 1`
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`D`

Show Worked Solution

`text(Use proper fraction tool on CAS:)`

`[text(CAS: propFrac) ((x – 3)/(2 – x))]`

`f(x) = -1 – 1/(x – 2)`

`:.\ text(Asymptotes:)\ \ x = 2, y = – 1`

`=>   D`