Question

The function f is such that f(x)=a^2x^2-ax+3b for x<1/(2a) Where a and b are constant for the case where a=1 and b=-1 Find f^-1(c.f. and find its domain I know domain of f^-1(x)=range of f(x) and it is -13/4 but I don't know inequality sign direction?

Answer 1

See below.

Explanation:

`a^2x^2-ax+3b`

`x^2-x-3`

Range:

Put into form `y=a(x-h)^2+k`

`h=-b/(2a)`

`k=f(h)`

`h=1/2`

`f(h)=f(1/2)=(1/2)^2-(1/2)-3=-13/4`

Minimum value `-13/4`

This occurs at `x=1/2`

So range is `(-13/4,oo)`

`f^(-1)(x)`

`x=y^2-y-3`

`y^2-y-(3-x)=0`

Using quadratic formula:

`y=(-(-1)+-sqrt((-1)^2-4(1)(-3-x)))/2`

`y=(1+-sqrt(4x+13))/2`

`f^(-1)(x)=(1+sqrt(4x+13))/2`

`f^(-1)(x)=(1-sqrt(4x+13))/2`

With a little thought we can see that for the domain we have the required inverse is:

`f^(-1)(x)=(1-sqrt(4x+13))/2`

With domain:

`(-13/4,oo)`

Notice that we had the restriction on the domain of `f(x)`

`x<1/2`

This is the x coordinate of the vertex and the range is to the left of this.