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

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

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

range

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

0 \leq f (x) \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 - x^{2} : - 2 \leq x \leq 2

Extreme Points of

4 - x^{2} :

Minimum

(- 2, 0),

Maximum

(0, 2),

Minimum

(2, 0)

Find the range for the interval

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

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

0 \leq f (x) \leq 2

- 6 y = 8 x + 1

x = - 7, (8, 5)

e^{- x}

f (x) = 4 x^{3} - 48 x - 8

(\frac{1}{6}, - \frac{1}{3}), (\frac{5}{6}, 5)

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