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

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

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extreme f(x)=5sin(5x)

extreme points

f (x) = 5 sin (5 x)

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

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

Plug the extreme point

x = \frac{π}{10} + \frac{2 π}{5} n

into

5 sin (5 x) \Rightarrow y = 5

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

Plug the extreme point

x = \frac{3 π}{10} + \frac{2 π}{5} n

into

5 sin (5 x) \Rightarrow y = - 5

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

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

f (x) = 4 x^{2} + 4 y = 16

\frac{3 x ^{2} - 2}{x ^{3} - 2 x - 8}

y = x^{2} - 4 x

\frac{x ^{3} - 1}{x ^{2}}

f (x) = ln (x^{2} - 1) + 1

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