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

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

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range of f(x)=sqrt(9-x^2)

range

f (x) = 9 - x^{2}

0 \leq f (x) \leq 3

Interval Notation

[0, 3]

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} :

Minimum

(- 3, 0),

Maximum

(0, 3),

Minimum

(3, 0)

Find the range for the interval

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

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

0 \leq f (x) \leq 3

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

\frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}}

\frac{3}{1 + 9 x ^{2}}

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