What is the range of sqrt(-4x^2+12) ?

The range of sqrt(-4x^2+12) is 0<= f(x)<= 2sqrt(3)

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range of sqrt(-4x^2+12)

range

- 4 x^{2} + 12

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

- 4 x^{2} + 12 : - 3 \leq x \leq 3

Simplify

- 4 x^{2} + 12 : 2 - x^{2} + 3

Extreme Points of

2 - x^{2} + 3 :

Minimum

(- 3, 0),

Maximum

(0, 23),

Minimum

(3, 0)

Find the range for the interval

- 3 \leq x \leq 3 : 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{x + 7}{x ^{2} - 16}

f (x) = 8 x^{3} = 2

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

- \frac{3}{2} cos (3 x - \frac{1}{2}) + 2

\frac{x - 1}{2 x + 1}

What is the range of sqrt(-4x^2+12) ?
The range of sqrt(-4x^2+12) is 0<= f(x)<= 2sqrt(3)