Question

What is the domain and range of `f(x)= sqrt(4x-x^2)`?

Answer 1

The domain is `x in [0,4]`
The range is `f(x) in [0,2]`

Explanation:

For the domain, what's under the square root sign is `>=0`

Therefore,

`4x-x^2>=0`

`x(4-x)>=0`

Let `g(x)=sqrt(x(4-x))`

We can build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `0` `color(white)(aaaaaa)` `4` `color(white)(aaaaaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `4-x` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `color(white)(aaa)` `+` `color(white)(aa)` `0` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `g(x)` `color(white)(aaaaaa)` `-` `color(white)(a)` `color(white)(aaa)` `0` `color(white)(aa)` `+` `color(white)(aa)` `0` `color(white)(aaaa)` `-`

Therefore

`g(x)>=0` when `x in [0,4]`

Let,

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

hen,

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

`x^2-4x+y^2=0`

The solutions this quadratic equation is when the discriminant `Delta>=0`

So,

`Delta=(-4)^2-4*1*y^2`

`16-4y^2>=0`

`4(4-y^2)>=0`

`4(2+y)(2-y)>=0`

Let `h(y)=(2+y)(2-y)`

We build the sign chart

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-2` `color(white)(aaaa)` #color(white)(aaaaaa)2# `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `2+y` `color(white)(aaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `2-y` `color(white)(aaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `h(y)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaaa)` `-`

Therefore,

`h(y)>=0`, when `y in [-2,2]`

This is not possible for the whole interval, so the range is `y in [0,2]`

graph{sqrt(4x-x^2) [-10, 10, -5, 5]}