What is the range of sqrt(16-x^2) ?

The range of sqrt(16-x^2) is 0<= f(x)<= 4

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range of sqrt(16-x^2)

range

16 - x^{2}

0 \leq f (x) \leq 4

Interval Notation

[0, 4]

Solution steps

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

Domain of

16 - x^{2} : - 4 \leq x \leq 4

Extreme Points of

16 - x^{2} :

Minimum

(- 4, 0),

Maximum

(0, 4),

Minimum

(4, 0)

Find the range for the interval

- 4 \leq x \leq 4 : 0 \leq f (x) \leq 4

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

0 \leq f (x) \leq 4

f (x) = \frac{( x - 5 )}{4}

f (x) = (x - 2)^{2} + 3

f (x) = \frac{x ^{2} + 3}{5 x - 2 x ^{2}}

f (x) = \frac{2 x - 5}{x ^{2} - 5 x + 6}

- \frac{1}{x + 4} - 1