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

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

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range of sqrt(64-x^2)

range

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

3 x + 4 y = - 12

2 x^{3} - 4 x^{2}

2 + x

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

f (x) = - \frac{3 x + 1}{11}