What is the range of f(x)=3-sqrt(4-x^2) ?

The range of f(x)=3-sqrt(4-x^2) is 1<= f(x)<= 3

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range of f(x)=3-sqrt(4-x^2)

range

f (x) = 3 - 4 - x^{2}

1 \leq f (x) \leq 3

Interval Notation

[1, 3]

Solution steps

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

Domain of

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

Extreme Points of

3 - 4 - x^{2} :

Maximum

(- 2, 3),

Minimum

(0, 1),

Maximum

(2, 3)

Find the range for the interval

- 2 \leq x \leq 2 : 1 \leq f (x) \leq 3

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

1 \leq f (x) \leq 3

(2, 3), (4, 8)

x (x - 6)^{2}

(- 1, 1), (- 2, - 1)

2 x^{2} + 7

f (x) = x^{4} - 8 x^{2} + 8

What is the range of f(x)=3-sqrt(4-x^2) ?
The range of f(x)=3-sqrt(4-x^2) is 1<= f(x)<= 3