How do you find the domain and range of #sqrt(25-x^2) #?

1 Answer
Jan 26, 2017

Answer:

The domain is #= [-5,5]#
The range is #=[0,5]#

Explanation:

What is under the #sqrt# sign is #>=0#

Therefore,

#25-x^2>=0#

#(5+x)(5-x)>=0#

Let #f(x)=(5+x)(5-x)#

We can now build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##5+x##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##5-x##color(white)(aaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#

Therefore,

#f(x)>=0# when #x in [-5, 5]#

The domain of #f(x)# is #D_f(x)= [-5,5]#

When #x=+-5#, #sqrt(f(x))=0#

When #x=0#, #sqrt(f(x))=5#, this is the max. value

so,
The range is #R = [0,5]#