Functions

State the domain and range for the functions represented by these two graphs.

a.

b. 

a. 

Domain: `x in RR` for `-1<=x<=5`

Range: `f(x) in RR` for `-8<=f(x)<=8` 

b.

Domain: `x in RR` for `-3<=x<=2`

Range: `f(x) in RR` for `-7<=f(x)<=20` 

Find the range for `f:x|->2x^2`  for `1<=x<=4`.

`f(x)=2x^2` is a quadratic function represented by a `uu` shaped parabola (positive coefficient)

Substitute `x=1`

 `f(1)=2(1)^2=2`

Substitute `x=4`

 `f(4)=2(4)^2=32`

Final solution

`2<=f(x)<=32`

Find the range for `f(x)=frac{12}{x}` for `1<=x<=8`.

`f(x)=frac{12}{x}` is a rational function represented by a hyperbola graph and `x!=0`

Substitute `x=1`

`f(1)=frac{12}{1}=12`

Substitute `x=8`

`f(8)=frac{12}{8}=1.5`

Final solution

`1.5<=f(x)<=12`

Find the range for `f(x)=1+sqrt{x-4}` for `x>=4`.

`f(x)=1+sqrt{x-4}` is a square root function represented by an increasing, upward-curve graph

Substitute `x=4`

`f(4)=1+sqrt{4-4}=1`

As x approaches infinity (no upper bound)

 `f(x)>=1`

The function `g:x|->5-ax-2x^2`, where a is a constant, is defined for `x in RR`. 

Find the range of g in terms of a.

The function represented by a `nn` shaped parabola (negative coefficient)

The vertex form of quadratic function `y=a(x-h)^2+k`, where `(h,k)` is the vertex of the parabola

`g(x)=-2x^2-ax+5`

Factorise

`g(x)=-2(x^2+frac{a}{2}x)+5`

The square of half the coefficient `frac{a}{2}`

`frac{a}{2}-:2=frac{a}{4}`

`(frac{a}{4})^2=frac{a^2}{16}`

Add and subtract to the equation

`g(x)=-2(x^2+frac{a}{2}x+frac{a^2}{16}-frac{a^2}{16})+5`

`g(x)=-2(x+frac{a}{4})^2+frac{a^2}{8}+5`

Vertex of the parabola

`(-frac{a}{4},frac{a^2}{8}+5)`

Maximum value of the function is `frac{a^2}{8}+5`

Final solution

`g(x)<=frac{a^2}{8}+5`

`f(x)=2x^2-8x+5` for `x in RR`, `0<=x<=k`

a. Express `f(x)` in the form `a(x+b)^2+c`.

b. State the value of `k` for which the graph of `y=f(x)` has a line of symmetry.

c. For your value of `k` from part b, find the range of `f`.

a. The function represented by a `uu` shaped parabola (positive coefficient)

Factorise

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

The square of half the coefficient `-4`

`-4-:2=-2`

`(-2)^2=4`

Add and subtract to the expression

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

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

Vertex of the parabola

`(2,-3)`

b. Vertical line `x=frac{-b}{2a}` that passes through the vertex is the line of symmetry

`x=-frac{(-8)}{2(2)}=2` (without domain restrictions)

For `f(x)=2x^2-8x+5` and `0<=x<=k`

`x=2` is the midpoint between `0` and `k`

Then `k=4`

c. Substitute `x=4`

`f(4)=2(4)^2-8(4)+5=5`

As the vertex of the parabola is when `f(2)=-3`

Final solution

`-3<=x<=5`

`f:x|->x^2` for `x in RR`

`g:x|->x+1` for `x in RR`

Express each of the following as a composite function, using only `f` and/or `g`

a. `x|->(x+1)^2`

b. `x|->x^2+2x+2`

c. `x|->x^4+2x^2+1`

a. `fg(x)` means the function `g` acts on `x` first, then `f` acts on the result

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

Final solution is `fg`

b. Rewrite `x^2+2x+2`

`x^2+2x+1+1`

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

For `(x+1)^2+1`, function `g(x)=x+1` must acts on `fg(x)=(x+1)^2`

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

Final solution is `gfg`

c. Rewrite `x^4+2x^2+1`

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

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

For `x^2+1`, function `f(x)=x^2` must acts on function `g(x)=x+1` 

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

For `(x^2+1)^2`, function `f(x)=x^2` must acts on `gf(x)=x^2+1`

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

Final solution is `fgf`

`f(x)=frac{x+5}{2x-1}` for `x in RR`, `x!=frac{1}{2}`

Show that `ff(x)=x`

`ff(x)=f(frac{x+5}{2x-1})`

`=frac{frac{x+5}{2x-1}+5}{2(frac{x+5}{2x-1})-1`

Multiply both the numerator and the denominator by `2x-1`

`frac{x+5+5(2x-1)}{2(x+5)-1(2x-1)}`

`=frac{11x}{11}`

 `=x`

`f:x|->(x+1)^3-4` for `x in RR`, `x>=0`

a. Find an expression for `f^-1(x)`.

b. Find the domain and range of `f^-1`.

a. `f(x)=(x+1)^3-4`

Write the function in terms of `y`

`y=(x+1)^3-4`

Interchange the `x` and `y` variables

`x=(y+1)^3-4`

Rearrange to make `y` the subject

 `x+4=(y+1)^3`

 `root(3)(x+4)=y+1`

 `y=root(3)(x+4)-1`

Final solution

 `f^-1(x)=root(3)(x+4)-1`

b. As function `f` is a cubic function, `f(x)` increases as `x` increases, and reaches `0` when `x=0` (minimum value)

Substitute `x=0`

`f(0)=(0+1)^3-4=-3`

The range of `f` is `f(x)>=-3` so the domain of `f^-1` is `x>=-3`

The domain of `f` is `x>=0` so the range of `f^-1(x)` is `f^-1(x)>=0`

`f(x)=frac{1}{x-1}` for `x in RR`, `x!=1`

a. Find an expression for `f^-1(x)`.

b. Show that if `f(x)=f^-1(x)`, then `x^2-x-1=0`.

c. Find the values of `x` for which `f(x)=f^-1(x)`. Give your answer in surd form.

a. Write the function in terms of `y`

`y=frac{1}{x-1}`

Interchange the `x` and `y` variables

`x=frac{1}{y-1}`

Rearrange to make `y` the subject

`x(y-1)=1`

`xy-x=1`

`xy=x+1`

`y=frac{x+1}{x}`

Final solution

`f^-1(x)=frac{x+1}{x}`

b. For `f(x)=f^-1(x)`

`frac{1}{x-1}=frac{x+1}{x}`

Multiply both sides by `x(x-1)`

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

`x=x^2-1`

`x^2-x-1=0`

c. For `x^2-x-1=0`

Using quadratic formula `x=frac{-b+-sqrt(b^2-4ac}}{2a}`

 `x=frac{-(-1)+-sqrt((-1)^2-4(1)(-1)}}{2(1)}`

 `x=frac{1+-sqrt5}{2}`

a. A cubic graph has equation `y=(x+3)(x-2)(x-5)`.

Write, in a similar form, the equation of the graph after a translation of `((2),(0))`.

b. The graph of `y=x^2-4x+1` is translated by the vector `((1),(2))`.

Find, in the form `y=ax^2+bx+c`, the equation of the resulting graph.

a. The graph of `y=f(x-a)` is a translation of the graph `y=f(x)` by the vector `((a),(0))`

After a translation of `((2),(0))`, the graph of `y=(x+3)(x-2)(x-5)` is replaced  all the `x`’s with `x-2`

`y=(x-2+3)(x-2-2)(x-2-5)`

`y=(x+1)(x-4)(x-7)`

b. The graph of `y=f(x-a)+b` is a translation of the graph `y=f(x)` by the vector `((a),(b))`

After a translation of `((1),(2))`, the graph of `y=x^2-4x+1` is replaced  all the `x`’s with `x-1`, and add `2` to the resulting function

`y=(x-1)^2-4(x-1)+1+2`

Rewrite

`y=x^2-2x+1-4x+4+1+2`

`y=x^2-6x+8`

The function `f` is defined by `f(x)=frac{2x}{3x-1}` for `x>frac{1}{3}`.

a. Find an expression for `f^-1(x)`.

b. Show that `frac{2}{3}+frac{2}{3(3x-1)}` can be expressed as `frac{2x}{3x-1}`.

c. State the range of `f`.

a. `y=frac{2x}{3x-1}`

Interchange the `x` and `y` variables

`x=frac{2y}{3y-1}`

Rearrange to make `y` the subject

`3xy-x=2y`

`y(3x-2)=x`

`y=frac{x}{3x-2}`

Final solution

`f^-1(x)=frac{x}{3x-2}`

b. The common denominator is `3(3x-1)`

`frac{2(3x-1)}{3(3x-1)}+frac{2}{3(3x-1)}`

`=frac{2(3x-1)+2}{3(3x-1)}`

`=frac{6x}{3(3x-1)}`

`=frac{2x}{3x-1}`

c. For `f(x)=frac{2x}{3x-1}`

`f(x)=frac{2}{3}+frac{2}{3(3x-1)}`

`f(x)=frac{2}{3}(1+frac{1}{3x-1})` (1)

And for `x>frac{1}{3}`

`3x>1`

`3x-1>0`

So `frac{1}{3x-1}` approach `0` as `x` increase

Substitute into (1)

`f(x)=frac{2}{3}(1+0)`

The range of `f` is `f(x)>frac{2}{3}`

Function `f` and `g` are defined by

`f(x)=4x-2` , for `x in RR`

`g(x)=frac{4}{x+1}` , for `x in RR`, `x!=-1`

a. Find the value of `fg(7)`.

b. Find the values of `x` for which `f^-1(x)=g^-1(x)`.

a. `g(7)=frac{4}{(7)+1}=frac{4}{8}=frac{1}{2}`

`fg(x)` means the function `g` acts on `x` first, then `f` acts on the result

`fg(7)=f(frac{1}{2})=4(frac{1}{2})-2=0`

b. For `f(x)=4x-2`

`y=4x-2`

Interchange the `x` and `y` variables

 `x=4y-2`

Rearrange to make `y` the subject

`4y=x+2`

`y=frac{x+2}{4}`

Then `f^-1(x)=frac{x+2}{4}`

For `g(x)=frac{4}{x+1}`

`y=frac{4}{x+1}`

Interchange the `x` and `y` variables

`x=frac{4}{y+1}`

Rearrange to make `y` the subject

`x(y+1)=4`

`y+1=frac{4}{x}`

`y=frac{4-x}{x}`

Then `g^-1(x)=frac{4-x}{x}`

`f^-1(x)=g^-1(x)`

 `frac{x+2}{4}=frac{4-x}{x}`

`x^2+2x=16-4x`

In each of parts a, b and c, the graph shown with solid lines has equation `y=f(x)`. The graph

shown with broken lines is a transformation of `y=f(x)`.

State, in terms of `f`, the equation of the graph shown with broken lines.

a.

b.

c.

a. The graph shown with broken lines is a reflection of the graph `y=f(x)` in the y-axis

Then `y=f(-x)`

b. The graph shown with broken lines is a stretch of `y=f(x)`, stretch factor `2`, parallel to the y-axis

Then `y=2f(x)`

c. 

The graph shown with broken lines is a translation of `y=f(x)` by the vector `((-4),(-3))`

Then `y=f(x+4)-3`

The diagram shows part of the graph of `y=atan(x-b)+c`.

Given that `0 < b < pi`, state the values of the constants a, b and c.

Center point O `(0,0)` from the original graph `y=tanx` is translated into point A `(frac{pi}{4},1)`

Using the tangent function `y=atan(x-b)+c`

`b=frac{pi}{4}` (the horizontal translation parallel to the x-axis)

`c=1` (the vertical translation of parallel to the y-axis)

For `y=atan(x-frac{pi}{4})+1`, substitute `x=0` and `y=-1`

`-1=atan(0-frac{pi}{4})+1`

`-2=a(-1)`

`a=2` (the amplitude)

Then `y=2tan(x-frac{pi}{4})+1`

The graph of `y=f(x)`  is transformed to the graph of `y=f(2x)-3`.

a. Describe fully the two single transformations that have been combined to give the resulting transformation.

b. The point P `(5,6)` lies on the transformed curve `y=f(2x)-3`. State the coordinates of the corresponding point on the original curve `y=f(x)`.

a. The graph of `y=f(2x)-3` is a stretch of `y=f(x)`, stretch factor `frac{1}{2}`, parallel to the x-axis, and followed by a translation by the vector `((0),(-3))`.

b. The point `(x,y)`  lies on the original curve `y=f(x)`

After transformation, the point `(x,y)`  changes to the new point `(2x,y+3)`

Substitute the point P `(5,6)` gives the coordinates `(2xx5,6+3)`

`=(10,9)`

The function `f` is defined by `f(x)=2x^2+3` for `x>=0`

a. Find and simplify an expression for `ff(x)`.

b. Solve the equation `ff(x)=34x^2+19`.

a. `ff(x)=2(2x^2+3)^2+3`

`=2(4x^4+12x^2+9)+3`

 `=8x^4+24x^2+18+3`

`=8x^4+24x^2+21`

b. For `ff(x)=34x^2+19`

`8x^4+24x^2+21=34x^2+19` 

`8x^4-10x^2+2=0`

`2(x^2-1)(4x^2-1)=0`

`x^2-1=0` or `4x^2-1=0`

`x^2=1` or `x^2=frac{1}{4}`

`x=+-1` or `x=+-frac{1}{2}`

Final solution

`x=1` or `x=frac{1}{2}` (as `x>=0`)

Functions `f` and `g` are defined as follows:

`f(x)=(x-2)^2-4` for `x>=2`

`g(x)=ax+2` for `x in RR`, where `a` is a constant.

a. State the range of `f`.

b. Find `f^-1(x)`.

c. Give that `a=-frac{5}{3}`, solve the equation `f(x)=g(x)`.

d. Given instead that `ggf^-1(12)=62`, find the possible values of `a`.

a. The function represented by a `uu` shaped parabola

Substitute `x=2`

`f(2)=(2-2)^2-4=-4`

Range: `f(x)>=-4`

b. For `f(x)=(x-2)^2-4`

`y=(x-2)^2-4`

Interchange the `x` and `y` variables

`x=(y-2)^2-4`

Rearrange to make `y` the subject

`(y-2)^2=x+4`

`y-2=+-sqrt(x+4)` 

`y=+-sqrt(x+4)+2`

`f^-1(x)=sqrt(x+4)+2` (as `x>=2`)

c. For `f(x)=g(x)`

`(x-2)^2-4=-frac{5}{3}x+2`

`x^2-4x+4-4=-frac{5}{3}x+2`

`x^2-frac{7}{3}x-2=0`

Using quadratic formula `x=frac{-b+-sqrt(b^2-4ac)}{2a}`

`x=frac{-(-frac{7}{3})+-sqrt((-frac{7}{3})^2-4(1)(-2))}{2(1)}`

`x=3` or `x=-frac{2}{3}`

Final solution

`x=3` (as `x>=2`)

d. For `f^-1(x)=sqrt(x+4)+2` and `g(x)=ax+2`

`f^-1(12)=sqrt(12+4)+2=6`

`gf^-1(12)=6a+2`

For `ggf^-1(12)=62`

`ggf^-1(12)=a(6a+2)+2=62`

`6a^2+2a+2-62=0`

`6a^2+2a-60=0`

`a=-frac{10}{3}` or `a=3`

Functions `f` and `g` are both defined for `x in RR`  and are given by

`f(x)=x^2-4x+9`

`g(x)=2x^2+4x+12`

a. Express `f(x)` in the form `(x-a)^2+b`.

b. Express `g(x)`  in the form `2[(x+c)^2+d]`.

c. Express `g(x)` in the form `kf(x+h)`, where `k`  and `h`  are integers.

d. Describe fully the two transformations that have been combined to transform the graph of `y=f(x)` to the graph of `y=g(x)`.

a. `f(x)=x^2-4x+9`

`=x^2-4x+4-4+9`

`=(x^2-4x+4)+5`

`=(x-2)^2+5`

b. `g(x)=2x^2+4x+12`

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

`=2(x^2+2x+1-1+6)`

`=2[(x^2+2x+1)-1+6]`

`=2[(x+1)^2+5]`

c. Substitute `x=x+3` into `f(x)=(x-2)^2+5`

`f(x+3)=(x+3-2)^2+5=(x+1)^2+5`

Then `g(x)=2[(x+1)^2+5]=2f(x+3)`

`k=2` and `h=3`

d. The graph of `g(x)=2f(x+3)` is a translation of the graph `y=f(x)` by the vector `((-3),(0))`

After a translation of `((-3),(0))`, the graph is stretched by factor `2` parallel to the y-axis.

`f(x)=(x+a)^2-a` for `x<=-a`, where `a`  is a positive constant

a. Find an expression for `f^-1(x)`.

b. State the domain and the range of the function `f^-1`.

a. For `f(x)=(x+a)^2-a`

`y=(x+a)^2-a`

Interchange the `x` and `y` variables

`x=(y+a)^2-a`

Rearrange to make `y` the subject

`(y+a)^2=x+a`

`y+a=+-sqrt(x+a)`

`y=+-sqrt(x+a)-a`

`f^-1(x)=-sqrt(x+a)-a` (as `x<=-a`)

b. Substitute `x=-a`

`f(-a)=(-a+a)^2-a=-a`

The range of `f` is `f(x)>=-a` so the domain of `f^-1` is `x>=-a`

The domain of `f` is `x<=-a` so the range of `f^-1(x)` is `f^-1(x)<=-a`