What is the range of-3/2 sin(2x-(3pi)/4)+7/3 ?

The range of-3/2 sin(2x-(3pi)/4)+7/3 is 5/6 <= f(x)<= 23/6

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range of-3/2 sin(2x-(3pi)/4)+7/3

range

- \frac{3}{2} sin (2 x - \frac{3 π}{4}) + \frac{7}{3}

\frac{5}{6} \leq f (x) \leq \frac{23}{6}

Interval Notation

[\frac{5}{6}, \frac{23}{6}]

Solution steps

The range of the basic

sin

function is

- 1 \leq sin (2 x - \frac{3 π}{4}) \leq 1

- 1 \leq sin (2 x - \frac{3 π}{4}) \leq 1

Multiply the edges of the range by:

- \frac{3}{2}

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

Add to the edges of the range:

\frac{7}{3}

\frac{5}{6} \leq - \frac{3}{2} sin (2 x - \frac{3 π}{4}) + \frac{7}{3} \leq \frac{23}{6}

Therefore the range is

\frac{5}{6} \leq f (x) \leq \frac{23}{6}

f (x) = (0, - 5) (5, 0)

f (x) = 2 (2)^{4} - 16 (2)^{3} + 8

(1, 3), (2, 5)

f (x) = - 12 x + 5

2.1 sin (3.8 t)

What is the range of-3/2 sin(2x-(3pi)/4)+7/3 ?
The range of-3/2 sin(2x-(3pi)/4)+7/3 is 5/6 <= f(x)<= 23/6