What is the range of sqrt(12-x^2) ?

The range of sqrt(12-x^2) is 0<= f(x)<= 2sqrt(3)

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range of sqrt(12-x^2)

range

12 - x^{2}

0 \leq f (x) \leq 23

Interval Notation

[0, 23]

Solution steps

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

Domain of

12 - x^{2} : - 23 \leq x \leq 23

Extreme Points of

12 - x^{2} :

Minimum

(- 23, 0),

Maximum

(0, 23),

Minimum

(23, 0)

Find the range for the interval

- 23 \leq x \leq 23 : 0 \leq f (x) \leq 23

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

0 \leq f (x) \leq 23

f (x) = \frac{- ( \frac{1}{5} ) x + 3}{5}

\frac{x - 2}{2 x ^{2}}

3 x + 6

f (x) = \frac{2 x - 3}{12 x ^{2} + 5 x - 3}

y = x^{2} - 9

What is the range of sqrt(12-x^2) ?
The range of sqrt(12-x^2) is 0<= f(x)<= 2sqrt(3)