What is the range of f(x)=sqrt(64-x^2) ?

The range of f(x)=sqrt(64-x^2) is 0<= f(x)<= 8

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range of f(x)=sqrt(64-x^2)

range

f (x) = 64 - x^{2}

0 \leq f (x) \leq 8

Interval Notation

[0, 8]

Solution steps

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

Domain of

64 - x^{2} : - 8 \leq x \leq 8

Extreme Points of

64 - x^{2} :

Minimum

(- 8, 0),

Maximum

(0, 8),

Minimum

(8, 0)

Find the range for the interval

- 8 \leq x \leq 8 : 0 \leq f (x) \leq 8

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

0 \leq f (x) \leq 8

(5, 1), (9, 4)

- \frac{3}{x + 2}

f (x) = 3 x^{4} + 12 x^{3}

f (x) = \frac{( 2 x + 1 )}{5}

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

What is the range of f(x)=sqrt(64-x^2) ?
The range of f(x)=sqrt(64-x^2) is 0<= f(x)<= 8