What is the range of sqrt(1-x^2) ?

The range of sqrt(1-x^2) is 0<= f(x)<= 1

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range of sqrt(1-x^2)

range

1 - x^{2}

0 \leq f (x) \leq 1

Interval Notation

[0, 1]

Solution steps

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

Domain of

1 - x^{2} : - 1 \leq x \leq 1

Extreme Points of

1 - x^{2} :

Minimum

(- 1, 0),

Maximum

(0, 1),

Minimum

(1, 0)

Find the range for the interval

- 1 \leq x \leq 1 : 0 \leq f (x) \leq 1

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

0 \leq f (x) \leq 1

f (x) = - x^{3} + 5 x^{2} + 8 x - 3

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

cos (3 x)

y - 70.75 = 0.11 (x - 250)

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