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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of y=sqrt(4-x^2)

range

y = 4 - x^{2}

0 \leq f (x) \leq 2

Interval Notation

[0, 2]

Solution steps

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

Domain of

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

Extreme Points of

4 - x^{2} :

Minimum

(- 2, 0),

Maximum

(0, 2),

Minimum

(2, 0)

Find the range for the interval

- 2 \leq x \leq 2 : 0 \leq f (x) \leq 2

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

0 \leq f (x) \leq 2

f (x) = \frac{5 - 3 x}{7 - 4 x}

sin (x)

f (x) = 3 csc (x + \frac{π}{2})

5 x + 6 y = 42

\frac{x ^{2} - 4}{7 x ^{2}}