What is the range of f(x)=sin^2(x) ?

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

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range of f(x)=sin^2(x)

range

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

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

sin^{2} (x) : - \infty < x < \infty

Extreme Points of

sin^{2} (x) :

Minimum

(πn, 0),

Maximum

(\frac{π}{2} + πn, 1)

Find the range for the interval

- \infty < x < \infty : 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) = 4 x

(7, 2), (7, 7)

f (x) = \frac{2}{3} x + 1

- \frac{1}{x - 1}

(100, 10500), (120, 11000)