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

1 Answer
Jul 31, 2017

Answer:

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]}