Question

Find the domain and range of the real valued function f(x) = 1/
$\sqrt{16 - {x}^{2} }$
​

Answer 1

Answer:

Domain : −4≤x≤4, in interval notation : [−4,4]

Range: 0≤f(x)≤4, in interval notation : [0,4]

Explanation:

f(x)=√16−x2 , for domain under root should not be

negative quantity. 16−x2≥0or16≥x2orx2≤16

∴x≤4orx≥−4 . Domain : −4≤x≤4or[−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)≤4or[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]}