What is the range of-sqrt(9-x^2) ?

The range of-sqrt(9-x^2) is -3<= f(x)<= 0

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range of-sqrt(9-x^2)

range

- 9 - x^{2}

- 3 \leq f (x) \leq 0

Interval Notation

[- 3, 0]

Solution steps

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

Domain of

- 9 - x^{2} : - 3 \leq x \leq 3

Extreme Points of

- 9 - x^{2} :

Maximum

(- 3, 0),

Minimum

(0, - 3),

Maximum

(3, 0)

Find the range for the interval

- 3 \leq x \leq 3 : - 3 \leq f (x) \leq 0

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

- 3 \leq f (x) \leq 0

(3.2) (- 11.3)

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

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

f (x) = 2 x - 8

G (t) = \frac{1 - 3 t}{4 + t}