What is the range of f(z)=sqrt(4-z^2) ?

The range of f(z)=sqrt(4-z^2) is 0<= f(z)<= 2

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range of f(z)=sqrt(4-z^2)

range

f (z) = 4 - z^{2}

0 \leq f (z) \leq 2

Interval Notation

[0, 2]

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

4 - z^{2} : - 2 \leq z \leq 2

Extreme Points of

4 - z^{2} :

Minimum

(- 2, 0),

Maximum

(0, 2),

Minimum

(2, 0)

Find the range for the interval

- 2 \leq z \leq 2 : 0 \leq f (z) \leq 2

Combine the ranges of all domain intervals to obtain the function range

0 \leq f (z) \leq 2

f (x) = x^{2} + x

f (x) = \frac{x - 1}{x ^{2} - 4 x + 3}

f (x) = (x + 3)^{3} - 1

m = - \frac{7}{16}

(x^{2} - 9)^{6}

What is the range of f(z)=sqrt(4-z^2) ?
The range of f(z)=sqrt(4-z^2) is 0<= f(z)<= 2