What is the extreme f(x)=sin(7x) ?

The extreme f(x)=sin(7x) is Maximum(pi/(14)+(2pi)/7 n,1),Minimum((3pi}{14}+\frac{2pi)/7 n,-1)

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extreme f(x)=sin(7x)

extreme points

f (x) = sin (7 x)

Maximum (\frac{π}{14} + \frac{2 π}{7} n, 1), Minimum (\frac{3 π}{14} + \frac{2 π}{7} n, - 1)

Solution steps

Find the critical points:

x = \frac{π}{14} + \frac{2 π}{7} n, x = \frac{3 π}{14} + \frac{2 π}{7} n

Domain of

sin (7 x) : - \infty < x < \infty

Find intervals:

Increasing

: \frac{2 π}{7} n \leq x < \frac{π}{14} + \frac{2 π}{7} n,

Decreasing

: \frac{π}{14} + \frac{2 π}{7} n < x < \frac{3 π}{14} + \frac{2 π}{7} n,

Increasing

: \frac{3 π}{14} + \frac{2 π}{7} n < x < \frac{2 π}{7} + \frac{2 π}{7} n

Plug the extreme point

x = \frac{π}{14} + \frac{2 π}{7} n

into

sin (7 x) \Rightarrow y = 1

Maximum (\frac{π}{14} + \frac{2 π}{7} n, 1)

Plug the extreme point

x = \frac{3 π}{14} + \frac{2 π}{7} n

into

sin (7 x) \Rightarrow y = - 1

Minimum (\frac{3 π}{14} + \frac{2 π}{7} n, - 1)

Maximum (\frac{π}{14} + \frac{2 π}{7} n, 1), Minimum (\frac{3 π}{14} + \frac{2 π}{7} n, - 1)

f (x) = \frac{7}{2} x - \frac{25}{2}

f (x) = x^{2} + 6 x + 4

y = \frac{7}{2} x - 2

f (x) = \frac{1 - 4 t}{6 + t}

\frac{3 x}{5 x - 3}

What is the extreme f(x)=sin(7x) ?
The extreme f(x)=sin(7x) is Maximum(pi/(14)+(2pi)/7 n,1),Minimum((3pi}{14}+\frac{2pi)/7 n,-1)