How do you find the domain and range of #sqrt(16-x^2)#?

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Answer 1 · 2017-08-06T11:36:33

Domain : # -4 <= x <= 4 #, in interval notation : #[-4,4]#
Range: # 0 <= f(x) <= 4 #, in interval notation : #[0,4]#

# f(x) = sqrt ( 16 -x^2) # , for domain under root should not be

negative quantity. # 16-x^2 >= 0 or 16 >= x^2 or x^2 <=16#

# :. x <= 4 or x >=-4# . Domain : # -4 <= x <= 4 or [-4,4]#

Range : # f(x) # is maximum at #x=0 , f(x) = 4# and

# f(x) # is minimum at #x=4 , f(x) = 0#

Range : # 0 <= f(x) <= 4 or [0,4]#

Domain : # -4 <= x <= 4 #, in interval notation : #[-4,4]#

Range: # 0 <= f(x) <= 4 #, in interval notation : #[0,4]#

graph{(16-x^2)^0.5 [-10, 10, -5, 5]}