What is the extreme f(x)=-3/2 sin(3/2 x) ?

The extreme f(x)=-3/2 sin(3/2 x) is Minimum(pi/3+(4pi)/3 n,-3/2),Maximum(pi+(4pi)/3 n, 3/2)

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extreme f(x)=-3/2 sin(3/2 x)

extreme

Minimum (\frac{π}{3} + \frac{4 π}{3} n, - \frac{3}{2}), Maximum (π + \frac{4 π}{3} n, \frac{3}{2})

Solution steps

Find the critical points:

x = \frac{π}{3} + \frac{4 π}{3} n, x = π + \frac{4 π}{3} n

Domain of

- \frac{3}{2} sin (\frac{3}{2} x) : - \infty < x < \infty

Find intervals:

Decreasing

: \frac{4 π}{3} n \leq x < \frac{π}{3} + \frac{4 π}{3} n,

Increasing

: \frac{π}{3} + \frac{4 π}{3} n < x < π + \frac{4 π}{3} n,

Decreasing

: π + \frac{4 π}{3} n < x < \frac{4 π}{3} + \frac{4 π}{3} n

Plug the extreme point

x = \frac{π}{3} + \frac{4 π}{3} n

into

- \frac{3}{2} sin (\frac{3}{2} x) \Rightarrow y = - \frac{3}{2}

Minimum (\frac{π}{3} + \frac{4 π}{3} n, - \frac{3}{2})

Plug the extreme point

x = π + \frac{4 π}{3} n

into

- \frac{3}{2} sin (\frac{3}{2} x) \Rightarrow y = \frac{3}{2}

Maximum (π + \frac{4 π}{3} n, \frac{3}{2})

Minimum (\frac{π}{3} + \frac{4 π}{3} n, - \frac{3}{2}), Maximum (π + \frac{4 π}{3} n, \frac{3}{2})

What is the extreme f(x)=-3/2 sin(3/2 x) ?
The extreme f(x)=-3/2 sin(3/2 x) is Minimum(pi/3+(4pi)/3 n,-3/2),Maximum(pi+(4pi)/3 n, 3/2)