square root of ( 16x⁴ +1 - 16x²)
Answers
Answer:
16 - x^2>=0 (otherwise the square root is not a real number)
x^2<=16
|x|<=4
-4<=x<=4
Domain=D=[-4;4]
It's obvious that
0<=16 - x^2<=16 in D ==>
0<=sqrt(16 - x^2)<=4 in D ==>
-2<=f(x)<=2 in D
Range=[-2;2]
To graph this function you need to make some transformations.
First let y=f(x) ==>
y = -2 + sqrt(16 - x^2)
y+2=sqrt(16 - x^2)
(y+2)^2=[sqrt(16 - x^2)]^2
(y+2)^2=16 - x^2
x^2+(y+2)^2=16
x^2+(y+2)^2=4^2
It's an equation of a circle with center at C(0;-2) and radius r=4.
But the graph of your function is only the upper half of that circle,
because Range=[-2;2]
F(x)=square root 64-x^2. what are domain and range?
Domain is the set of values which you can put into the function. F(x) is a square root function which will not be defined for negative values inside the root so you need to solve
64 - x^2 >= 0
--> (8-x)(8+x) >=0
--> -8 <= x <= 8
--> [-8, 8]
Range is the set of outputs from the function. From our domain F(x) is at its maximum when x = 0 (F(0) = sqr(64 - 0^2) = 8) and F(x) is at its minimum when x = -8 or x = 8 (F(8) = 0)
So the range is [0,8]
Step-by-step explanation: