What is the range of sin(x)cos(x) ?

The range of sin(x)cos(x) is -1/2 <= f(x)<= 1/2

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range of sin(x)cos(x)

range

sin (x) cos (x)

- \frac{1}{2} \leq f (x) \leq \frac{1}{2}

Interval Notation

[- \frac{1}{2}, \frac{1}{2}]

Solution steps

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

Domain of

sin (x) cos (x) : - \infty < x < \infty

Extreme Points of

sin (x) cos (x) :

Maximum

(\frac{π}{4} + πn, \frac{1}{2}),

Minimum

(\frac{3 π}{4} + πn, - \frac{1}{2})

Find the range for the interval

- \infty < x < \infty : - \frac{1}{2} \leq f (x) \leq \frac{1}{2}

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

- \frac{1}{2} \leq f (x) \leq \frac{1}{2}

\frac{x ^{2} - 5 x + 4}{x + 1}

f (x) = \frac{6}{1 + 0.8 e ^{- 2 x}}

\frac{x ^{n} + 2}{x ^{2 n} - 4}

\frac{e ^{x} - e ^{- x}}{3}

f (x) = \frac{e ^{x}}{6 + e ^{x}}

What is the range of sin(x)cos(x) ?
The range of sin(x)cos(x) is -1/2 <= f(x)<= 1/2