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Functions, 2ADV F1 EQ-Bank 11

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`.

  1. Find  `g(f(3))`.  (2 marks)

  2. 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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  1. `2`
  2. `-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 SM-Bank 31

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< ∞`

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`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 SM-Bank 30

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

  1. Find  `g(f(x))`.  (1 mark)

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

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  1. `25 – x`
  2. `text(Domain:)\ −5<= x <= 5`

     

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

Show Worked Solution
i.    `g(f(x))` `= 25 – (f(x))^2`
    `= 25 – (sqrtx)^2`
    `= 25 – x`

 

ii.    `f(g(x))` `= sqrt(g(x))`
    `= sqrt(25 – x^2)`

 
`:.\ text(Domain:)\ −5<= x <= 5`

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

Functions, 2ADV F1 SM-Bank 12 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`

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`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 SM-Bank 8 MC

Let  `f (x) = x^2`

Which one of the following is not true?

A.   `f(xy) = f (x) f (y)`

B.   `f(x) - f(-x) = 0`

C.   `f (2x) = 4 f (x)`

D.   `f (x - y) = f(x) - f(y)`

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`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 SM-Bank 7

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

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

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

  3. Show that  `h(x) + h(−x) = f ((g(x))^2 )`.  (2 marks)
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  1. `log_e(x^2 + 1)`
  2. `text(Domain)\ (h):\ text(all)\ x`

     

    `text(Range)\ h(x):\ \  h>=0`

  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `h(x)` `= f(x^2 + 1)`
    `= log_e(x^2 + 1)`

 

ii.   `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.
iii.   `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 SM-Bank 5 MC

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

Then  `f(g(x))`  is given by
 

A.   `e^(4x + 6) + 2 e^(2x + 3) - 3`

B.   `2x^2 + 4x - 6`

C.   `e^(2x^2 + 4x - 3)`

D.   `e^(2x^2 + 4x - 6)`

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`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`