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

The extreme f(x)=sin((pix)/2) is Maximum(1+4n,1),Minimum(3+4n,-1)

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extreme f(x)=sin((pix)/2)

extreme points

f (x) = sin (\frac{π x}{2})

Maximum (1 + 4 n, 1), Minimum (3 + 4 n, - 1)

Solution steps

Find the critical points:

x = 1 + 4 n, x = 3 + 4 n

Domain of

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

Find intervals:

Increasing

: 4 n \leq x < 1 + 4 n,

Decreasing

: 1 + 4 n < x < 3 + 4 n,

Increasing

: 3 + 4 n < x < 4 + 4 n

Plug the extreme point

x = 1 + 4 n

into

sin (\frac{π x}{2}) \Rightarrow y = 1

Maximum (1 + 4 n, 1)

Plug the extreme point

x = 3 + 4 n

into

sin (\frac{π x}{2}) \Rightarrow y = - 1

Minimum (3 + 4 n, - 1)

Maximum (1 + 4 n, 1), Minimum (3 + 4 n, - 1)

- 0.2 x^{3} - 0.7 x^{2} + 3 x - 6

h (t) = \frac{t}{t - 2} [3.5]

\frac{d}{d t} ((r^{3} - 2 r^{2} + 6 t + 2) 3)

f (x) = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 10

f (x) = - 9 - x^{2}

What is the extreme f(x)=sin((pix)/2) ?
The extreme f(x)=sin((pix)/2) is Maximum(1+4n,1),Minimum(3+4n,-1)