What is the range of 4sin(2x-pi/3) ?

The range of 4sin(2x-pi/3) is -4<= f(x)<= 4

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range of 4sin(2x-pi/3)

range

4 sin (2 x - \frac{π}{3})

- 4 \leq f (x) \leq 4

Interval Notation

[- 4, 4]

Solution steps

The range of the basic

sin

function is

- 1 \leq sin (2 x - \frac{π}{3}) \leq 1

- 1 \leq sin (2 x - \frac{π}{3}) \leq 1

Multiply the edges of the range by:

4

- 4 \leq 4 sin (2 x - \frac{π}{3}) \leq 4

Therefore the range is

- 4 \leq f (x) \leq 4

y = x^{2} + 7

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

y = 5^{\frac{x}{3}}

16 x - 8 y = 17

f (x) = - 5 sin (\frac{3}{2} x)

What is the range of 4sin(2x-pi/3) ?
The range of 4sin(2x-pi/3) is -4<= f(x)<= 4