What is the range of f(x)=sqrt(16-x^2) ?

The range of f(x)=sqrt(16-x^2) is 0<= f(x)<= 4

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range of f(x)=sqrt(16-x^2)

range

f (x) = 16 - x^{2}

0 \leq f (x) \leq 4

Interval Notation

[0, 4]

Solution steps

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

Domain of

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

Extreme Points of

16 - x^{2} :

Minimum

(- 4, 0),

Maximum

(0, 4),

Minimum

(4, 0)

Find the range for the interval

- 4 \leq x \leq 4 : 0 \leq f (x) \leq 4

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

0 \leq f (x) \leq 4

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

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

cos (x - \frac{π}{2})

f (x) = 2 x^{2} + 1

6 x - 7 y + 11 = 0

What is the range of f(x)=sqrt(16-x^2) ?
The range of f(x)=sqrt(16-x^2) is 0<= f(x)<= 4