Question

What is the domain and range of `f(x)={ x^2 - 81 }/ {x^2 - 4x}`?

Answer 1

`D_f=RR-{0,4}=(-oo,0)uu(0,4)uu(4,+oo)` , Range = `f(D_f)=(-oo,(81-9sqrt65)/8]uu[(81+9sqrt65)/8,+oo)`

Explanation:

`f(x)=(x^2-81)/(x^2-4x)`

In order for this function to be defined we need `x^2-4x!=0`

We have `x^2-4x=0` `<=>` `x(x-4)=0` `<=>` `(x=0,x=4)`

So `D_f=RR-{0,4}=(-oo,0)uu(0,4)uu(4,+oo)`

For `x` `inD_f` ,

`f(x)=(x^2-81)/(x^2-4x)` `=` `((x-9)(x+9))/(x^2-4x)`

`f(x)=0 <=> (x=9,x=-9)`

• `(x^2-81)/(x^2-4x)=y` `<=>` `x^2-81=y(x^2-4x)`

`x^2-81=yx^2-4xy`

• Adding `color(green)(4yx)` in both sides,

`x^2-81+4yx=yx^2`

• Substracting `color(red)(yx^2)` from both sides

`x^2-81+4yx-yx^2=0` `<=>`

`x^2(1-y)+4xy-81=0`

This is quadratic equation for `x` so
`a=1-y`
`b=4y`
`c=-81`

We need `D=b^2-4*a*c>=0` `<=>`
`16y^2-4(1-y)*(-81)>=0` `<=>`
`16y^2+324(1-y)>=0` `<=>`
`16y^2-324y+324>=0` `<=>`
`4y^2-81y+81>=0`

`y_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)`

`=` `(81+-sqrt(6561-1296))/8`

`=` `(81+-sqrt(5265))/8`

`=` `(81+-9sqrt65)/8`

`4y^2-81y+81>=0` `<=>` `(y<=(81-9sqrt65)/8` or `y>=(81+9sqrt65)/8)`

so, `f(x)<=(81-9sqrt65)/8` or `f(x)>=(81+9sqrt65)/8`

Which means, `f(D_f)=(-oo,(81-9sqrt65)/8]uu[(81+9sqrt65)/8,+oo)`