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

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

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extreme f(x)=sin(2pix)

extreme points

f (x) = sin (2 π x)

Maximum (\frac{1}{4} + 1 \cdot n, 1), Minimum (\frac{3}{4} + 1 \cdot n, - 1)

Solution steps

Find the critical points:

x = \frac{1}{4} + 1 \cdot n, x = \frac{3}{4} + 1 \cdot n

Domain of

sin (2 π x) : - \infty < x < \infty

Find intervals:

Increasing

: 1 \cdot n \leq x < \frac{1}{4} + 1 \cdot n,

Decreasing

: \frac{1}{4} + 1 \cdot n < x < \frac{3}{4} + 1 \cdot n,

Increasing

: \frac{3}{4} + 1 \cdot n < x < 1 + 1 \cdot n

Plug the extreme point

x = \frac{1}{4} + 1 n

into

sin (2 π x) \Rightarrow y = 1

Maximum (\frac{1}{4} + 1 \cdot n, 1)

Plug the extreme point

x = \frac{3}{4} + 1 n

into

sin (2 π x) \Rightarrow y = - 1

Minimum (\frac{3}{4} + 1 \cdot n, - 1)

Maximum (\frac{1}{4} + 1 \cdot n, 1), Minimum (\frac{3}{4} + 1 \cdot n, - 1)

f (x) = x^{2} + y + x y

y = 2 + 3 x - x^{3}

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

6 x y - x^{2} y - x y^{2}

f (x) = 2 x^{3} - 3 x^{2} - 72 x + 9, - 4 \leq x \leq 5

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