What is the extreme f(x)=-8sin(4x) ?

The extreme f(x)=-8sin(4x) is Minimum(pi/8+pi/2 n,-8),Maximum((3pi)/8+pi/2 n,8)

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extreme f(x)=-8sin(4x)

extreme points

f (x) = - 8 sin (4 x)

Minimum (\frac{π}{8} + \frac{π}{2} n, - 8), Maximum (\frac{3 π}{8} + \frac{π}{2} n, 8)

Solution steps

Find the critical points:

x = \frac{π}{8} + \frac{π}{2} n, x = \frac{3 π}{8} + \frac{π}{2} n

Domain of

- 8 sin (4 x) : - \infty < x < \infty

Find intervals:

Decreasing

: \frac{π}{2} n \leq x < \frac{π}{8} + \frac{π}{2} n,

Increasing

: \frac{π}{8} + \frac{π}{2} n < x < \frac{3 π}{8} + \frac{π}{2} n,

Decreasing

: \frac{3 π}{8} + \frac{π}{2} n < x < \frac{π}{2} + \frac{π}{2} n

Plug the extreme point

x = \frac{π}{8} + \frac{π}{2} n

into

- 8 sin (4 x) \Rightarrow y = - 8

Minimum (\frac{π}{8} + \frac{π}{2} n, - 8)

Plug the extreme point

x = \frac{3 π}{8} + \frac{π}{2} n

into

- 8 sin (4 x) \Rightarrow y = 8

Maximum (\frac{3 π}{8} + \frac{π}{2} n, 8)

Minimum (\frac{π}{8} + \frac{π}{2} n, - 8), Maximum (\frac{3 π}{8} + \frac{π}{2} n, 8)

f (x) = 16 - 4 x - \frac{4}{x}, [0, \infty]

f (x) = - 7 x - 6

f (x) = \frac{x ^{2}}{1 - x ^{2}}

f (x) = - 6 x^{2} + 84 x + 1000

f (x) = \frac{4 x}{x ^{2} + 1}, - 3 \leq x \leq 3

What is the extreme f(x)=-8sin(4x) ?
The extreme f(x)=-8sin(4x) is Minimum(pi/8+pi/2 n,-8),Maximum((3pi)/8+pi/2 n,8)