What is the extreme f(x)=sin(5t) ?

The extreme f(x)=sin(5t) is Maximum(pi/(10)+(2pi)/5 n,1),Minimum((3pi}{10}+\frac{2pi)/5 n,-1)

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extreme f(x)=sin(5t)

extreme points

f (x) = sin (5 t)

Maximum (\frac{π}{10} + \frac{2 π}{5} n, 1), Minimum (\frac{3 π}{10} + \frac{2 π}{5} n, - 1)

Solution steps

Find the critical points:

t = \frac{π}{10} + \frac{2 π}{5} n, t = \frac{3 π}{10} + \frac{2 π}{5} n

Domain of

sin (5 t) : - \infty < t < \infty

Find intervals:

Increasing

: \frac{2 π}{5} n \leq t < \frac{π}{10} + \frac{2 π}{5} n,

Decreasing

: \frac{π}{10} + \frac{2 π}{5} n < t < \frac{3 π}{10} + \frac{2 π}{5} n,

Increasing

: \frac{3 π}{10} + \frac{2 π}{5} n < t < \frac{2 π}{5} + \frac{2 π}{5} n

Plug the extreme point

x = \frac{π}{10} + \frac{2 π}{5} n

into

sin (5 t) \Rightarrow y = 1

Maximum (\frac{π}{10} + \frac{2 π}{5} n, 1)

Plug the extreme point

x = \frac{3 π}{10} + \frac{2 π}{5} n

into

sin (5 t) \Rightarrow y = - 1

Minimum (\frac{3 π}{10} + \frac{2 π}{5} n, - 1)

Maximum (\frac{π}{10} + \frac{2 π}{5} n, 1), Minimum (\frac{3 π}{10} + \frac{2 π}{5} n, - 1)

y = - \frac{2}{5}, (- 2, 1)

f (x) = \frac{4 - x}{x - 16}

y + 3 = x

f (x) = 2 - 4 - x

f (x) = 6 x + 5

What is the extreme f(x)=sin(5t) ?
The extreme f(x)=sin(5t) is Maximum(pi/(10)+(2pi)/5 n,1),Minimum((3pi}{10}+\frac{2pi)/5 n,-1)