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

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

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range of f(x)=sqrt(1-x^2)

range

f (x) = 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

y = x^{2} + 1

f (x) = x^{2} tan (x)

\frac{20}{( x + 5 ) ^{2}}

\frac{3 x ^{2} - 12 x + 13}{x ^{2} - 4 x + 4}

y = - 125 x + 850

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